Rd Sharma XII Vol 1 2019 Solutions for Class 12 Science Math Chapter 4 Inverse Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Inverse Trigonometric Functions are extremely popular among Class 12 Science students for Math Inverse Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2019 Book of Class 12 Science Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2019 Solutions. All Rd Sharma XII Vol 1 2019 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 4.10:

Question 1:

Find the domain of definition of fx=cos-1x2-4.

Answer:

For cos-1x2-4 to be defined
-1x2-413x25x-5,-33,5
Hence, the domain of fx is -5,-33,5.

Page No 4.10:

Question 2:

Find the domain of fx=2cos-12x+sin-1x.

Answer:

For 2cos-12x to be defined.
    -12x1-12x12    .....i
For sin-1x to be defined.
    -1x1      .....ii
Domain of fx=-12, 12-1, 1
                       =-12,12.

Page No 4.10:

Question 3:

Find the domain of fx=cos-1x+cosx.

Answer:

For cos-1x to be defined.
-1x1
Now, cosx is defined for all real values.
So, domain of cosx is R.
Domain of fx is R-1, 1=-1, 1.

Page No 4.10:

Question 4:

​Find the principal values of each of the following:

(i) cos-1-32
(ii) cos-1-12
(iii) cos-1sin4π3

(iv) cos-1tan3π4

Answer:

(i)  Let cos-1-32=y
Then, 
cosy=-32
We know that the range of the principal value branch is 0,π.
Thus, 
cosy=-32=cos5π6y=5π60,π
Hence, the principal value of cos-1-32 is 5π6.

(ii) Let cos-1-12=y
Then, 
cosy=-12
We know that the range of the principal value branch is 0, π.
Thus, 
cosy=-12=cos3π4y=3π40,π
Hence, the principal value of cos-1-12 is 3π4.

(iii) Let cos-1sin4π3=y
Then, 
cosy=sin4π3
We know that the range of the principal value branch is 0,π.
Thus, 
cosy=sin4π3=-32=cos5π6y=5π60,π
Hence, the principal value of cos-1sin4π3 is 5π6.

(iv) Let cos-1tan3π4=y
Then, 
cosy=tan3π4
We know that the range of the principal value branch is 0,π.
Thus, 
cosy=tan3π4=-1=cosπy=π0,π
Hence, the principal value of cos-1tan3π4 is π.

Page No 4.10:

Question 5:

For the principal values, evaluate each of the following:

(i)  cos-112+2sin-112
(ii) 
(iii) sin-1-12+2 cos-1-32
(iv) sin-1-32+cos-132

Answer:

 (i) cos-1cosx=x sin-1sinx=x

cos-112+2sin-112= cos-1cosπ3+2sin-1sinπ6=π3+2π6= 2π3

(ii) 

 iii sin-1-12+2cos-1-32=sin-1sin-π6+2cos-1cos5π6      Range of sine is -π2, π2 ;  -π6 -π2, π2 and range of cosine is 0, π ; 5π6 0, π=-π6+25π6=-π6+5π3=9π6=3π2
sin-1-12+2cos-1-32=3π2

iv sin-1-32+cos-132=sin-1sin-π3+cos-1cosπ6         =-π3+π6                 Range of sine is -π2, π2 ;  -π3 -π2, π2 and range of cosine is 0, π ; π6 0, π=-π6
sin-1-32+cos-132=-π6



 



Page No 4.115:

Question 1:

Evaluate the following:

(i) 
tan2 tan-115-π4
(ii) tan12cos-153
(iii) sin12cos-145
(iv) sin2tan-123+costan-13

Answer:

(i)

tan2 tan-115-π4=tan2 tan-115-tan-1 1                              =tantan-12×151-152-tan-1 1  2 tan-1x=tan-12x1-x2                             =tantan-1252425-tan-1 1                             =tantan-1512+tan-1 1                             =tantan-1512-11+512          tan-1x-tan-1y=tan-1x-y1+xy                             =tantan-1-7121712                             =tantan-1-717                             =-717

(ii)
 Let, cos-153=θcosθ=532cos2θ2-1=53cos2θ2=3+56cosθ2=3+56θ2=cos-13+56        =tan-11-3+5623+56        =tan-11-3+563+56        =tan-13-563+56        =tan-13-53+5        =tan-13-53-53+53-5        =tan-13-529-5        =tan-13-52i.e., 12cos-153=tan-13-52tan12cos-153=tantan-13-52tan12cos-153=3-52

(iii)
sin12cos-145=sin12×2sin-1±1-452    cos-1x=2sin-1±1-x2                        =sinsin-1±110                        =±110

(iv) 
sin2tan-123+costan-13=sinsin-12×231+49+coscos-111+32=sinsin-11213+coscos-112=1213+12=3726

Page No 4.115:

Question 2:

(i) 2sin-135=tan-1247
(ii) tan-114+tan-129=12cos-135=12sin-145
(iii)   tan-123=12tan-1125

(iv) tan-117+2 tan-113=π4

(v) sin-145+2 tan-113=π2
(vi) 2 sin-135-tan-11731=π4
(vii) 2 tan-115+tan-118=tan-147
(viii) 2 tan-134-tan-11731=π4
(ix)  2 tan-112+tan-117=tan-13117
(x) 4tan-115-tan-1 1239=π4

Answer:

 i LHS=2 sin-135              =2 tan-1341-925       sin-1x=tan-1x1-x2              =2 tan-13545               =2 tan-134              =tan-12×341-342  2 tan-1x=tan-12x1-x2              =tan-132716              =tan-1=RHS                             


ii LHS=tan-114+tan-1 29          =tan-114+291-14×29          tan-1x+tan-1y=tan-1x+y1-xy          =tan-117363436          =tan-112          =12cos-11-141+14                   tan-1x=12cos-11-x21+x2          =12cos-13454          =12cos-135Now,tan-112=12sin-1221+14    tan-1x=12sin-12x1+x2                  =12sin-1154                  =12sin-145                             

iii LHS=tan-123           =12tan-12×231-232  tan-1x=12tan-12x1-x2           =12tan-14359           =12tan-1125=RHS                             

iv LHS=tan-117+2tan-1 13           =tan-117+tan-12×131-132  2 tan-1x=tan-12x1-x2           = tan-117+tan-12389           = tan-117+tan-134          =tan-117+341-17×34          tan-1x+tan-1y=tan-1x+y1-xy          =tan-125282528          =tan-11=π4=RHS                             

v LHS=sin-145+2tan-1 13           =sin-145+tan-12×131-132  2 tan-1x=tan-12x1-x2           = sin-145+tan-12389           = sin-145+tan-134           = sin-145+cos-111+916      tan-1x=cos-111+x2           = sin-145+cos-1154           = sin-145+cos-145           = π2=RHS                             

vi LHS=2 sin-135-tan-1 1731              =2 tan-1341-925-tan-1 1731       sin-1x=tan-1x1-x2              =2 tan-13545-tan-1 1731               =2 tan-134-tan-1 1731               =tan-12×341-342-tan-1 1731  2 tan-1x=tan-12x1-x2              =tan-132716-tan-1 1731              =tan-1247-tan-1 1731              =tan-1247-17311+247×1731          tan-1x-tan-1y=tan-1x-y1+xy              =tan-1625217625217              =tan-11=π4=RHS                             

vii LHS=2 tan-115+tan-1 18                              =tan-12×151-152+tan-1 18  2 tan-1x=tan-12x1-x2                             =tan-1252425+tan-1 18                             =tan-1512+tan-1 18                           =tan-1512+181-512×18          tan-1x+tan-1y=tan-1x+y1-xy                             =tan-113249196                             =tan-147=RHS                             

viii LHS=2 tan-134-tan-1 1731                              =tan-12×341-342-tan-1 1731  2 tan-1x=tan-12x1-x2                             =tan-132716-tan-1 1731                             =tan-1247-tan-1 1731                           =tan-1247-17311+247×1731          tan-1x-tan-1y=tan-1x-y1+xy                             =tan-1625217625217                             =tan-11=π4=RHS                             

ix LHS=2 tan-112+tan-1 17                              =tan-12×121-122+tan-1 17  2 tan-1x=tan-12x1-x2                             =tan-1134+tan-1 17                             =tan-143+tan-1 17                           =tan-143+171-43×17          tan-1x+tan-1y=tan-1x+y1-xy                             =tan-131211721                             =tan-13117=RHS                             

x LHS=4tan-115-tan-1 1239               =2tan-12×151-152-tan-1 1239  2 tan-1x= tan-12x1-x2              =2tan-1252425-tan-1 1239              =2tan-1512-tan-1 1239              =tan-12×5121-5122-tan-1 1239  2 tan-1x= tan-12x1-x2              =tan-156119144-tan-1 1239              =tan-1120119-tan-1 1239              =tan-1120119-172391+120119×1239          tan-1x- tan-1y=tan-1x-y1+xy              =tan-11=π4=RHS                             

 

Page No 4.115:

Question 3:

If sin-12a1+a2-cos-11-b21+b2=tan-12x1-x2, then prove that x=a-b1+ab

Answer:

Let: a=tanm b=tann x=tany

Now,
sin-12a1+a2-cos-11-b21+b2=tan-12x1-x2sin-12tanm1+tan2m-cos-11-tan2n1+tan2n=tan-12tany1-tan2ysin-1sin2m-cos-1cos2n=tan-1tan2y     sin2x=2tanx1+tan2x and cos2x=1-tan2x1+tan2x2m-2n=2ym-n=ytan-1a-tan-1b=tan-1x      a=tanm, b=tann and x=tanytan-1a-b1+ab=tan-1x          tan-1x-tan-1y=tan-1x-y1+xya-b1+ab=x

a-b1+ab=x

Page No 4.115:

Question 4:

Prove that
(i) tan-11-x22x+cot-11-x22x=π2

(ii) sintan-11-x22x+cos-11-x22x=1
 

Answer:

(i)
tan-11-x22x+cot-11-x22x=π2LHS=tan-11-x22x+cot-11-x22x        =tan-11-x22x+π2-tan-11-x22x     tan-1x+cot-1x=π2        =π2=RHS

(ii)
 sintan-11-x22x+cos-11-x21+x2=1LHS=sintan-11-x22x+cos-11-x21+x2        =sinsin-11-x22x1+1-x22x+cos-11-x21+x2              tan-1x=sin-1x1+x2        =sinsin-11-x21+x2+cos-11-x21+x2        =sinπ2                 sin-1x+cos-1x=π2                =1=RHS

Page No 4.115:

Question 5:

If sin-12a1+a2+sin-12b1+b2=2 tan-1 x, prove that x=a+b1-ab.

Answer:

Let:
a=tanzb=tany

Then,
sin-12a1+a2+sin-12b1+b2=2tan-1xsin-12tanz1+tan2z+sin-12tany1+tan2y=2tan-1xsin-1sin2z+sin-1sin2y=2tan-1x       sin2x=2tanx1+tan2x2z+2y=2tan-1xtan-1a+tan-1b=tan-1x        a=tanz and b=tanytan-1a+b1-ab=tan-1x          tan-1x+tan-1y=tan-1x+y1-xy     x=a+b1-ab

Page No 4.115:

Question 6:

Show that 2 tan−1 x + sin−1 2x1+x2 is constant for x ≥ 1, find that constant.

Answer:

We have
2tan-1x+sin-12x1+x21 For x>1,=2tan-1x+sin-12x1+x2=π-sin-12x1+x2+sin-12x1+x2         2tan-1x=π-sin-12x1+x2 ,  x>1=π2 For x=1,=2tan-1x+sin-12x1+x2   =2tan-11+sin-1211+12=2tan-11+sin-11=2π4+π2  =π



Page No 4.116:

Question 7:

Find the values of each of the following:
(i) tan-12 cos2 sin-112
(ii)  cossec-1x+cosec-1x,  x 1

Answer:

(i) Let  sin-112=y
Then,
siny=12

 tan-12cos2sin-112=tan-12cos2y=tan-121-2sin2y        cos2x=1-2sin2x=tan-121-2×14           siny=12=tan-12×12=tan-11=π4

 tan-12cos2sin-112=π4
(ii)
We have
cossec-1x+cosec-1x=cosπ2    sec-1x+cosec-1x=π2=0

 cossec-1x+cosec-1x=0  , |x|1

Page No 4.116:

Question 8:

Solve the following equations for x:

(i) 
tan-114+2 tan-115+tan-116+tan-11x=π4

(ii) 3 sin-12x1+x2-4 cos-11-x21+x2+2 tan-12x1-x2=π3

(iii)  tan-12x1-x2+cot-11-x22x=2π3, x>0

(iv) 2 tan-1 (sinx=tan-1 (2 sinx), xπ2.

(v)cos-1x2-1x2+1+12tan-12x1-x2=2π3

(vi) tan-1 x-2x-1+tan-1 x+2x+1=π4

Answer:

(i) We know
 tan-1x+tan-1y=tan-1x+y1-xy

 tan-114+2tan-115+tan-116+tan-11x=π4tan-114+tan-115+tan-115+tan-116+tan-11x=π4tan-114+151-14×15+tan-115+161-15×16+tan-11x=π4tan-19201920+tan-111302930+tan-11x=π4tan-1919+tan-11129+tan-11x=π4tan-1919+11291-1129×919+tan-11x=π4tan-1235226+tan-11x=π4tan-1235226+1x1-235226×1x=π4235x+226226x-235=tanπ4235x+226226x-235=1235x+226=226x-2359x=-461x=-4619

(ii)

 3sin-12x1+x2-4cos-11-x21+x2+2tan-12x1-x2=π36tan-1x-8tan-1x+4tan-1x=π3        2tan-1x=sin-12x1+x2, 2tan-1x=cos-11-x21+x2 and 2tan-1x=tan-12x1-x2     2tan-1x=π3tan-1x=π6    x=tanπ6x=13

(iii) We know
tan-1x+tan-1y=tan-1x+y1-xy

 tan-12x1-x2+cot-11-x22x=2π3tan-12x1-x2+tan-12x1-x2=2π3        cot-1x=tan-11xtan-12x1-x2=π32tan-1x=π3             2tan-1x tan-12x1-x2tan-1x=π6x=tanπ6x=13

(iv) 2 tan-1 (sinx=tan-1 (2 sinx), xπ2

tan-12sinx1-sin2x=tan-12sinx             2tan-1x=tan-12x1-x22sinx1-sin2x=2sinx2sinx=2sinx-2sin3x2sin3x=0sinx=0x=0

(v) cos-1x2-1x2+1+12tan-12x1-x2=2π3cos-11-x21+x2+12×2tan-1x=2π3                           tan-12x1-x2=2tan-1x2tan-1x+tan-1x=2π3                                                 cos-11-x21+x2=2tan-1x3tan-1x=2π3tan-1x=2π9x=tan2π9

(vi)
tan-1x-2x-1+tan-1x+2x+1=π4tan-1x-2x-1+tan-1x+2x+1=tan-11tan-1x-2x-1=tan-11-tan-1x+2x+1tan-1x-2x-1=tan-11-x+2x+11+x+2x+1tan-1x-2x-1=tan-1x+1-x-2x+1+x+2tan-1x-2x-1=tan-1-12x+3x-2x-1=-12x+32x2+3x-4x-6=-x+12x2=1+6x2=7x=±72

Page No 4.116:

Question 9:

Prove that 2 tan-1a-ba+btanθ2=cos-1a cos θ+ba+b cos θ

Answer:

LHS=2 tan-1a-ba+btanθ2=cos-11-a-ba+btanθ221+a-ba+btanθ22     2 tan-1x=cos-11-x21+x2                                                       =cos-11-a-ba+btan2θ21+a-ba+btan2θ2                                                       =cos-1a+b-a-btan2θ2a+b+a-btan2θ2                                                       =cos-1a+b-atan2θ2+btan2θ2a+b+atan2θ2-btan2θ2                                                       =cos-1a1-tan2θ2+b1+tan2θ2a1+tan2θ2+b1-tan2θ2                                                    =cos-1a1-tan2θ21+tan2θ2+b1+tan2θ21+tan2θ2a1+tan2θ21+tan2θ2+b1-tan2θ21-tan2θ2    Dividing Nr and Dr by 1+tan2θ2                                                    =cos-1a1-tan2θ21+tan2θ2+ba+b1-tan2θ21-tan2θ2                                                    =cos-1acosθ+ba+bcosθ=RHS

Page No 4.116:

Question 10:

Prove that:
tan-12aba2-b2+tan-12xyx2-y2=tan-12αβα2-β2,
where α = axby and β = ay + bx.

Answer:

We know
tan-1x+tan-1y=tan-1x+y1-xy,  xy>1

 tan-12aba2-b2+tan-12xyx2-y2=tan-12aba2-b2+2xyx2-y21-2aba2-b22xyx2-y2=tan-12abx2-aby2+xya2-xyb2a2-b2x2-y2a2x2-a2y2-x2b2+y2b2-4abxya2-b2x2-y2=tan-12abx2-aby2+xya2-xyb2a2x2-a2y2-x2b2+y2b2-2abxy-2abxy=tan-12ax-byay+bxax-by2-ay+bx2=tan-12αβα2-β2           α=ax-by and β=ay+bx

Page No 4.116:

Question 11:

For any a, b, x, y > 0, prove that:
23tan-13ab2-a3b3-3a2b+23tan-13xy2-x3y3-3x2y=tan-12αβα2-β2
where α = − ax + by, β = bx + ay

Answer:

Let a=btanm and x=ytann
Then,

23tan-13ab2-a3b3-3a2b+23tan-13xy2-x3y3-3x2y=23tan-13b3tanm-b3tan3mb3-3b3tan2m+23tan-13y3tann-y3tan3ny3-3y3tan2n=23tan-13tanm-tan3m1-3tan2m+23tan-13tann-tan3n1-tan2n=23tan-1tan3m+23tan-1tan3n         tan3x=3tanx-tan3x1-3tan2x=233m+233n =2m+2n=2tan-1ab+tan-1xy       a=btanm, x=ytann=2tan-1ab+xy1-abxy=2tan-1ay+bxby-ax=tan-12ay+bxby-ax1-ay+bxby-ax2=tan-12ay+bxby-axby-ax2-ay+bx2=tan-12αβα2-β2       β=ay+bx and α=by-ax

Page No 4.116:

Question 1:

Write the value of sin-1-32+cos-1-12

Answer:

 sin-1-x=-sin-1x, x-1,1cos-1-x=π-cos-1x,  x-1,1


 sin-1-32+cos-1-12=-sin-132+π-cos-112 = -sin-1sinπ3+π-cos-1cosπ3=-π3+π-π3=π3
  sin-1-32+cos-1-12=π3

Page No 4.116:

Question 2:

Write the difference between maximum and minimum values of sin−1 x for x ∈ [− 1, 1].

Answer:

The maximum value of sin-1x in x-1,1 is at 1.
So, the maximum value is

sin-11 = sin-1sinπ2=π2

Again, the minimum value is at -1.
Thus, the minimum value is

sin-1-1=-sin-11 = -sin-1π2=-π2
So, the difference between the maximum and the minimum value is
 π2--π2=π

Page No 4.116:

Question 3:

If sin−1 x + sin−1 y + sin−1 z = 3π2, then write the value of x + y + z.

Answer:

sin-1x+sin-1y+sin-1z=3π2sin-1x+sin-1y+sin-1z=π2+π2+π2        As the maximum value in the range of sin-1x is π2And here sum of three inverse of sine is 3 times π2. i.e., every sin inverse function is equal to π2 here.sin-1x=π2, sin-1y=π2 and sin-1z=π2x=1, y=1 and z=1  x+y+z=1+1+1=3



Page No 4.117:

Question 4:

If x > 1, then write the value of sin12x1+x2 in terms of tan−1 x.

Answer:

sin-12x1+x2=π-2tan-1x         2tan-1x=π- sin-12x1+x2 for x>1

Page No 4.117:

Question 5:

If x < 0, then write the value of cos−11-x21+x2 in terms of tan−1 x.

Answer:

Let x=tany
Then,
cos-11-x21+x2=cos-11-tan2y1+tan2y=cos-1cos2y      1-tan2x1+tan2x=cos2x=2y                  ...1

The value of x is negative.
So, let x = -a where a > 0.

-a=tanyy=tan-1-a
Now,

cos-11-x21+x2=2y         Using 1=2tan-1-a           =-2tan-1x          x=-a

Page No 4.117:

Question 6:

Write the value of tan1x + tan−11x for x > 0.

Answer:

tan-1x+tan-1y=tan-1x+y1-xy, xy<1

 tan-1x+tan-11x=tan-1x+1x1-x1x,x>0=tan-1x2+10=tan-1=tan-1tanπ2=π2

 tan-1x+tan-11x=π2

Page No 4.117:

Question 7:

Write the value of tan1 x + tan−1 1x for x < 0.

Answer:

 tan-1x+tan-1y=tan-1x+y1-xy
When x<0, 1x<0, then both are negative.
Let x = - y, y>0
Then,
tan-1x+tan-11x=tan-1-y+tan-1-1y=-tan-1y+tan-11y=-tan-1y+1y1-y1y,y>0=-tan-1y2+10=-tan-1=-tan-1tanπ2=-π2
 tan-1x+tan-11x=-π2, x<0

Page No 4.117:

Question 8:

What is the value of cos−1cos2π3+sin-1sin2π3?

Answer:

cos-1cos 2π3+sin-1sin 2π3=cos-1cos 2π3+sin-1sin π-π3   =cos-1cos 2π3+sin-1sin π3                       Range of sine is -π2, π2 ;  π3 -π2, π2  and range of cosine is 0, π ;  2π3 0, π       =2π3+π3=π

Page No 4.117:

Question 9:

If −1 < x < 0, then write the value of sin-12x1+x2+cos-11-x21+x2.

Answer:

Let x=-tany 
where 0<y<π2
Then,
sin-12x1+x2+cos-11-x21+x2=sin-1-2tany1+tan2y+cos-11-tan2y1+tan2y=sin-1-sin 2y+cos-1cos2y =-sin-1sin 2y+cos-1cos2y =-2y+2y     =0

 sin-12x1+x2+cos-11-x21+x2=0

Page No 4.117:

Question 10:

Write the value of sin (cot−1 x).

Answer:

We know
cot-1x=tan-11x
Now, we have
sincot-1x=sintan-11x=sinsin-11x1+1x2        tan-1x=sin-1x1+x2=sinsin-11xx2+1x=sinsin-1 1x2+1=1x2+1      sinsin-1x=x

Hence, sincot-1x=1x2-1 

Page No 4.117:

Question 11:

Write the value of cos-112+2 sin-112.

Answer:

We have
cos-112+2sin-112=cos-1cosπ3+2sin-1sinπ6              The range of sine is -π2, π2;  π6 -π2, π2  and the range of cosine is 0, π;  π3 0, π       =π3+2π6=π3+π3=2π3
cos-112+2sin-112=2π3

Page No 4.117:

Question 12:

Write the range of tan−1 x.

Answer:

The range of  tan-1x is-π2,π2.

Page No 4.117:

Question 13:

Write the value of cos−1 (cos 1540°).

Answer:

We know that 
cos-1cosx=x
Now,
cos-1cos1540=cos-1cos1440+100=cos-1cos100       cos4π+100=cos100°=100

Page No 4.117:

Question 14:

Write the value of sin−1 sin(-600°).

Answer:

We know that sin-1sinx=x.
Now,

sin-1sin-600=sin-1sin720-600=sin-1sin120=sin-1sin180-120        sinx = sinπ-x=sin-1sin60=60

∴ sin-1sin-600=60

Page No 4.117:

Question 15:

Write the value of cos 2 sin-113.

Answer:

Let y=sin-113
Then, siny=13

Now, cosy=1-sin2y,

cosy=1-19=89=223

cos2sin-113=cos(2y)=cos2y-sin2y     cos 2x= cos2x-sin2x=2232-132=89-19=79

∴ cos2sin-113=79

Page No 4.117:

Question 16:

Write the value of sin1 (sin 1550°).

Answer:

We know that sin-1sinx=x.
Now,
sin-1sin1550=sin-1sin1620-1550        sinx =sin1620-x=sin-1sin70=70

∴ sin-1sin1550=70

Page No 4.117:

Question 17:

Evaluate sin 12cos-145.

Answer:

We know that
cos-1x=2tan-11-x1+xtan-1x=sin-1x1+x2

 sin12cos-145=sin122tan-11-451+45=sintan-11595=sintan-113=sinsin-1131+19=sinsin-1110=110      sinsin-1x=x

∴ sin12cos-145=110

Page No 4.117:

Question 18:

Evaluate sin tan-134.

Answer:

We know that
tan-1x=sin-1x1+x2

 sintan-134=sinsin-1341+916=sinsin-13454=sinsin-135=35      sinsin-1x=x

​∴ sintan-134=35

Page No 4.117:

Question 19:

Write the value of cos−1 tan3π4.

Answer:

We have
cos-1tan3π4=cos-1-tanπ-3π4      tanπ-x=-tanx=cos-1tan-π4=cos-1-tanπ4=cos-1-1=cos-1cosπ       cosπ=-1 =π

∴ cos-1tan3π4=π

Page No 4.117:

Question 20:

Write the value of cos 2 sin-112.

Answer:

We have, cos 2 sin-112 = cos2×π6=cosπ3=12

Page No 4.117:

Question 21:

Write the value of cos1 (cos 350°) − sin−1 (sin 350°)

Answer:

We have
cos-1cos350-sin-1sin350=cos-1cos360-350-sin-1sin360-350         sin360-x=-sinx ,  cos360-x=cosx                                                          =cos-1cos10-sin-1sin-10=10--10=20

∴ cos-1cos350-sin-1sin350=20

Page No 4.117:

Question 22:

Write the value of cos2 12cos-135.

Answer:

Let y=cos-135cosy=35
Now,
cos212cos-135=cos212y=cosy+12    cos2x = 2cos2x-1=35+12=852=45
cos212cos-135=45

Page No 4.117:

Question 23:

If tan−1 x + tan−1 y = π4, then write the value of x + y + xy.

Answer:

We know that tan-1x+tan-1y=tan-1x+y1-xy.
Now,
tan-1x+tan-1y=π4tan-1x+y1-xy=π4x+y1-xy=tanπ4x+y1-xy=1   x+y=1-xyx+y+xy=1

∴ x+y+xy=1

Page No 4.117:

Question 24:

Write the value of cos−1 (cos 6).

Answer:

We know that cos-1cosx=x.
Now,
cos-1cos6=cos-1cos2π-6=2π-6  

Page No 4.117:

Question 25:

Write the value of sin−1 cosπ9.

Answer:

Consider,
sin-1cosπ9=sin-1sinπ2-π9   cos x =sinπ2-x=sin-1sin7π18=7π18        sin-1sinx  =x
  
∴ sin-1cosπ9=7π18 

Page No 4.117:

Question 26:

Write the value of sin π3-sin-1-12.

Answer:

We have
sinπ3-sin-1-12=sinπ3--π6=sinπ3+π6=sinπ2=1

∴ sinπ3-sin-1-12 = 1

Page No 4.117:

Question 27:

Write the value of tan1 tan15π4.

Answer:

We have
tan-1tan15π4=tan-1tan4π-π4tan-1-tanπ4    tan4π-x=-tanx=tan-1tan-π4      =-π4    tan-1tanx =x  


∴  tan-1tan15π4=-π4



Page No 4.118:

Question 28:

Write the value of 2sin-112+cos-1-12 .

Answer:

2sin-112+cos-1-12=sin-12×121-122+cos-1-12=sin-132+cos-1-12=sin-1sinπ3+cos-1cos2π3=π3+2π3=π 

Page No 4.118:

Question 29:

Write the value of tan-1ab-tan-1a-ba+b.

Answer:

We know that tan-1x-tan-1y=tan-1x-y1+xy.
Now,
tan-1ab-tan-1a-ba+b=tan-1ab-a-ba+b1+aba-ba+b=tan-1a2+ab-ab+b2ba+bab+b2-ab+a2ba+b=tan-11=tan-1tanπ4      tanπ4=1=π4

∴ tan-1ab-tan-1a-ba+b=π4

Page No 4.118:

Question 30:

Write the value of cos−1 cos5π4.

Answer:

 cos-1cos5π45π4 as 5π4 does not lie between 0 and π.
We have

cos-1cos5π4=cos-1cos2π-3π4=cos-1cos3π4=3π4

Page No 4.118:

Question 31:

Show that sin-1(2x1-x2)=2 sin-1x.

Answer:

We have
LHS=sin-12x1-x2Putting x=sin a, we get=sin-12 sin a1-sin2a      =sin-12sin a cos a=sin-1sin 2a=2a=2sin-1x    x=sin a

Page No 4.118:

Question 32:

Evaluate: sin-1sin3π5.

Answer:

We know that  sin-1sinx=x.
We have
sin-1sin3π5=sin-1sinπ-3π5    π-3π5-π2,π2=sin-1sin2π5=2π5

∴ sin-1sin3π5=2π5

Page No 4.118:

Question 33:

If tan-1(3)+cot-1x=π2, find x.

Answer:

We know that tan-1x+cot-1x=π2.
We have
tan-13+cot-1x=π2tan-13=π2-cot-1xtan-13=tan-1xx=3

x=3

Page No 4.118:

Question 34:

If sin-113+cos-1x=π2, then find x.

Answer:

We know that sin-1x+cos-1x=π2.
We have
sin-113+cos-1x=π2sin-113=π2-cos-1xsin-113=sin-1xx=13

∴ x=13

Page No 4.118:

Question 35:

Write the value of sin-113-cos-1-13.

Answer:

We know that sin-1x+cos-1x=π2 and cos-1-x=π-cos-1x.

 sin-113-cos-1-13=sin-113-π-cos-113=sin-113-π+cos-113=sin-113+cos-113-π=π2-π      sin-1x+cos-1x=π2=-π2

∴ sin-113-cos-1-13=-π2

Page No 4.118:

Question 36:

If 4 sin−1 x + cos−1 x = π, then what is the value of x?

Answer:

We know that sin-1x+cos-1x=π2 

 4sin-1x+cos-1x=π4sin-1x+π2-sin-1x=π    sin-1x+cos-1x=π23sin-1x=π 2sin-1x=π 6x=sinπ 6x=12

∴ x=12

Page No 4.118:

Question 37:

If x < 0, y < 0 such that xy = 1, then write the value of tan1 x + tan−1 y.

Answer:

We know
 tan-1x+tan-1y=tan-1x+y1-xy

 x<0, y<0 such that 
xy=1

Let x = -a and y = -b where both a and b are positive.

 tan-1x+tan-1y=tan-1x+y1-xy=tan-1-a-a1-1=tan-1-=tan-1tan-π2=-π2

Page No 4.118:

Question 38:

What is the principal value of sin-1-32?

Answer:

Let y=sin-1-32
Then,
siny=-32=sin-π3y=-π3-π2,π2

Here, -π2,π2 is the range of the principal value branch of inverse sine function.

∴ sin-1-32=-π3

Page No 4.118:

Question 39:

Write the principal value of sin-1-12

Answer:

Let y=sin-1-12
Then,
siny=-12=sin-π6y=-π6-π2,π2

Here, -π2,π2 is the range of the principal value branch of the inverse sine function.

∴ sin-1-12=-π6

Page No 4.118:

Question 40:

Write the principal value of cos-1cos2π3+sin-1sin2π3

Answer:

We have, cos-1cos2π3+sin-1sin2π3=cos-1cos2π3+sin-1sinπ-π3     π-2π3-π2,π2=cos-1cos2π3+sin-1sinπ3=2π3+π3=π

∴ cos-1cos2π3+sin-1sin2π3=π

Page No 4.118:

Question 41:

Write the value of tan2tan-115

Answer:

tan2tan-115=tantan-12×151-152=tantan-1252425=tantan-1512=512

Page No 4.118:

Question 42:

Write.the principal value of tan-11+cos-1-12

Answer:

tan-11+cos-1-12=tan-1tanπ4+cos-1cos2π3=π4+2π3=11π12

Page No 4.118:

Question 43:

Write the value of tan-12sin2cos-132

Answer:

tan-12sin2cos-132=tan-12sincos-12322-1=tan-12sincos-112

Page No 4.118:

Question 44:

Write the principal value of tan-13+cot-13

Answer:

We know tan-1x+cot-1x=π2
tan-13+cot-13=π2

Page No 4.118:

Question 45:

Write the principal value of cos-1cos680°

Answer:

cos-1cos680°=cos-1cos720°-680°=cos-1cos40°=40°

Page No 4.118:

Question 46:

Write the value of sin-1sin3π5

Answer:

sin-1sin3π5=sin-1sinπ-2π5=sin-1sin2π5=2π5

Page No 4.118:

Question 47:

Write the value of sec-112.

Answer:

The value of sec-112 is undefined as it is outside the range i.e., R – (–1, 1) .

Page No 4.118:

Question 48:

Write the value of cos-1cos14π3

Answer:

cos-1cos14π3=cos-1cos4π+2π3=cos-1cos2π3=2π3

Page No 4.118:

Question 49:

Write the value of cossin-1x+cos-1x, x1

Answer:

We have
x1±x1x1 or -x1x1 or x-1x-1, 1
Now,
cossin-1x+cos-1x=cosπ2       sin-1x+cos-1x=π2=0

Page No 4.118:

Question 50:

Wnte the value of the expression tansin-1x+cos-1x2, when x=32

Answer:

tansin-1x+cos-1x2=tanπ4       sin-1x+cos-1x=π2=1



Page No 4.119:

Question 51:

Write the principal value of sin-1cossin-112

Answer:

sin-1cossin-112=sin-1cossin-1sinπ3=sin-1cosπ3=sin-112=sin-1sinπ3=π3

Page No 4.119:

Question 52:

The set of values of cosec-132

Answer:

The value of cosec-132 is undefined as it is outside the range i.e., R – (–1, 1) .

Page No 4.119:

Question 53:

Write the value of  tan-11x for x < 0 in terms of cot-1x

Answer:

tan-11x=tan-1-1x      for x<0=-tan-11x=-cot-1x=-π-cot-1x=-π+cot-1x

Page No 4.119:

Question 54:

Write the value of  cot-1-x for all xR  in terms of cot-1x

Answer:

We know that cot-1-x=π-cot-1x
Therefore, the value of  cot-1-x for all xR  in terms of cot-1x is π-cot-1x.

Page No 4.119:

Question 55:

Wnte the value of  costan-1x+cot-1x3, when x=-13

Answer:

costan-1x+cot-1x3=cosπ6       tan-1x+cot-1x=π2=32

Page No 4.119:

Question 56:

If costan-1x+cot-13=0, find the value of x.

Answer:

costan-1x+cot-13=0costan-1x+cot-13=cosπ2tan-1x+cot-13=π2x=3        tan-1y+cot-1y=π2

Page No 4.119:

Question 57:

Find the value of 2sec-12+sin-112

Answer:

2sec-12+sin-112=2sec-1secπ3+sin-1sinπ6=2×π3+π6=5π6

Page No 4.119:

Question 58:

If cossin-125+cos-1x=0, find the value of x.

Answer:

cossin-125+cos-1x=0cossin-125+cos-1x=cosπ2sin-125+cos-1x=π2x=25             sin-1y+cos-1y=π2

Page No 4.119:

Question 59:

Find the value of cos-1cos13π6

Answer:

cos-1cos13π6=cos-1cos2π+π6=cos-1cosπ6=π6

Page No 4.119:

Question 60:

Find the value of tan-1tan9π8

Answer:

tan-1tan9π8=tan-1tanπ+π8=tan-1tanπ8=π8

Page No 4.119:

Question 61:

Find the value of tan-13-cot-1-3.

Answer:


tan-13-cot-1-3=tan-1tanπ3-cot-1cot5π6       Range of tan-1 is -π2, π2 ;  π3  -π2, π2 and range of cot-1 is 0, π ; 5π6 0, π=π3-5π6=-π2

Page No 4.119:

Question 1:

If tan-11+x2-1-x21+x2+1-x2 = α, then x2 =
(a) sin 2 α
(b) sin α
(c) cos 2 α
(d) cos α

Answer:

(a) sin 2α

tan-11+x2-1-x21+x2+1-x2=α1+x2-1-x21+x2+1-x2=tanα  1+x2-1-x21+x2+1-x2×1+x2-1-x21+x2-1-x2  =tanα1+x22+1-x22-21+x21-x21+x22-1-x22=tanα1-1-x4x2=tanαx2tanα=1-1-x41-x4=1-x2tanα1-x4=1+x4tan2α-2x2tanαx4+x4tan2α-2x2tanα=0x4sec2α-2x2tanα=0x2x2sec2α-2tanα=0x2sec2α-2tanα=0       x20x2sec2α=2tanαx2=2tanαsec2α=2sinαcosα=sin2α



Page No 4.120:

Question 2:

The value of tan cos-1152-sin-1417 is
(a) 293

(b) 293

(c) 329

(d) 329

Answer:

(d) 329

Let, cos-1152=y and sin-1417=z

 cosy=152siny=752tany=7sinz=417cosz=117tanz=4

 tancos-1152-sin-1417=tany-z=tany-tanz1+tany tanz=7-41+7×4=329

Page No 4.120:

Question 3:

2 tan−1 {cosec (tan−1 x) − tan (cot1 x)} is equal to
(a) cot−1 x

(b) cot−11x

(c) tan−1 x

(d) none of these

Answer:

(c) tan−1 x

Let tan-1x=y
So, x=tany

 2tan-1cosectan-1x-tancot-1x=2tan-1cosectan-1x-tantan-11x    =2tan-1cosectan-1x-1x=2tan-1cosec y-1tany=2tan-11-cosysiny=2tan-12sin2y2siny  =2tan-12sin2y22siny2cosy2=2tan-1tany2=y=tan-1x      

Page No 4.120:

Question 4:

If cos-1xa+cos-1yb=α, thenx2a2-2xyabcos α+y2b2=
(a) sin2 α
(b) cos2 α
(c) tan2 α
(d) cot2 α

Answer:

(a) sin2 α

We know that cos-1x+cos-1y=cos-1xy-1-x21-y2.

 cos-1xa+cos-1yb=αcos-1xayb-1-x2a21-y2b2=αxyab-1-x2a21-y2b2=cosα1-x2a21-y2b2=xyab-cosα1-x2a21-y2b2=x2a2y2b2+cos2α-2xyabcosα     Squaring both the sides1-x2a2-y2b2+x2a2y2b2=x2a2y2b2+cos2α-2xyabcosαx2a2+y2b2-2xyabcosα=1-cos2α=sin2α

Page No 4.120:

Question 5:

The positive integral solution of the equation
tan-1x+cos-1y1+y2=sin-1310is
(a) x = 1, y = 2
(b) x = 2, y = 1
(c) x = 3, y = 2
(d) x = −2, y = −1.

Answer:

(a) x = 1, y = 2

We have,tan-1x+cos-1y1+y2=sin-1310tan-1x+tan-11-y1+y22y1+y2=tan-13101-3102tan-1x+tan-11y=tan-13tan-1x+1y1-x×1y=tan-13xy+1y-x=33y-3x=xy+13x+xy=3y-1x3+y=3y-1x=3y-13+yFor, y=1 x=12For, y=2 x=1For, y=3 x=43For, y=4 x=117For, y=1 x=73 and so on......Therefore, only integral solutions are :x=1 and y=2

Page No 4.120:

Question 6:

If sin−1 x − cos−1 x = π6, then x =
(a) 12

(b) 32

(c) -12

(d) none of these

Answer:

(b) 32
We know that sin-1x+cos-1x=π2.

 sin-1x-cos-1x=π6π2-cos-1x -cos-1x=π6-2cos-1x =π6-π2-2cos-1x =-π3cos-1x =π6x=cosπ6x=32

Page No 4.120:

Question 7:

sin cot-1tancos-1x is equal to
(a) x

(b) 1-x2

(c) 1x

(d) none of these

Answer:

(a) x

Let cos-1x=y

Then,
sincot-1tancos-1x=sincot-1tan y =sincot-1cot π2-y   =sinπ2-y=cosy     =x         cosy=x

Page No 4.120:

Question 8:

The number of solutions of the equation
tan-12x+tan-13x=π4 is
(a) 2
(b) 3
(c) 1
(d) none of these

Answer:

(a) 2
We know that tan-1x+tan-1y=tan-1x+y1-xy.

 tan-12x+tan-13x=π4tan-12x+3x1-2x ×3x=π42x+3x1-2x ×3x=tanπ45x1-6x2=1  5x=1-6x26x2+5x-1=0

Therefore, there are two solutions.

Page No 4.120:

Question 9:

If α = tan-1tan5π4 and β=tan-1-tan2π3, then
(a) 4 α = 3 β
(b) 3 α = 4 β
(c) α − β = 7π12
(d) none of these

Answer:

(a) 4 α = 3 β

We know that tan-1tanx=x.

 α=tan-1tan5π4=tan-1tanπ+π4=tan-1tanπ4=π4
and
β=tan-1-tan2π3=tan-1-tanπ-π3=tan-1tanπ3=π3

 4α=π3β=π
∴ 4α = 3β

Page No 4.120:

Question 10:

The number of real solutions of the equation
1+cos 2x=2sin-1(sin x),-πxπ is
(a) 0
(b) 1
(c) 2
(d) infinite

Answer:

(c) 2

For, -πx-π21+cos 2x=2sin-1(sin x)2 cos x=2 -π-x2 -cos x=2 -π-xcosx=π+x It does not satisfy for any value of x in the interval -π, -π2

For, -π2xπ21+cos 2x=2sin-1(sin x)2 cos x=2 x2 cos x=2 xcosx= x It gives one value of x in the interval -π2, π2

For, π2xπ1+cos 2x=2sin-1(sin x)2 cos x=2 -π-x2 -cos x=2 π-xcosx=-π+x It gives one value of x in the interval π2, π

1+cos 2x=2sin-1(sin x) gives two real solutions in the interval -π, π

Page No 4.120:

Question 11:

If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
(a) π2

(b) -π2

(c) − π

(d) none of these

Answer:

(b) -π2
We know that tan-1x+tan-1y=tan-1x+y1-xy.
 x<0, y<0 such that
 xy=1

Let x = -a and y = -b, where a and b both are positive.

 tan-1x+tan-1y=tan-1x+y1-xy=tan-1-a-a1-1=tan-1-=tan-1tan-π2=-π2

Page No 4.120:

Question 12:

If u=cot-1tan θ-tan-1tan θthen, tanπ4-u2=
(a) tan θ

(b) cot θ

(c) tan θ

(d) cot θ

Answer:

(a) tan θ

Let y=tanθ
Then,
u=cot-1tanθ-tan-1tanθu=cot-1y-tan-1yu=π2-2tan-1y        tan-1x+cot-1x=π2 2tan-1y=π2-u tan-1y=π4-u2y=tanπ4-u2tanθ=tanπ4-u2  y=tanθ

Page No 4.120:

Question 13:

If cos-1x3+cos-1y2=θ2, then 4x2-12xy cosθ2+9y2=
(a) 36
(b) 36 − 36 cos θ
(c) 18 − 18 cos θ
(d) 18 + 18 cos θ

Answer:

(c) 18 − 18 cosθ

We know
 cos-1x+cos-1y=cos-1xy-1-x21-y2

 cos-1x3+cos-1y2=θ2cos-1x3y2-1-x291-y24=θ2xy6-9-x294-y24=cosθ2xy-6cosθ2=9-x2 4-y2

Squaring both the sides, we get
x2y2-12xycosθ2+36cos2θ2=9-x24-y2x2y2-12xycosθ2+36cos2θ2=36-9y2-4x2+x2y24x2+9y2-12xycos2θ2=36-36cos2θ24x2+9y2-12xycos2θ2=361-cosθ+12    cos2x =2cos2x-14x2+9y2-12xycos2θ2=18-18cosθ

Page No 4.120:

Question 14:

If α = tan-13x2y-x, β=tan-12x-y3y, then α − β =
(a) π6

(b) π3

(c) π2

(d) -π3

Answer:

(a) π6


We have
α = tan-13x2y-x, β=tan-12x-y3y
Now, α-β=tan-13x2y-x-tan-12x-y3y                =tan-13x2y-x-2x-y3y1+3x2y-x×2x-y3y                =tan-13xy-4xy+2y2+2x2-xy3y2y-x3y2y-x+3x2x-y3y2y-x                =tan-13xy-4xy+2y2+2x2-xy23y2-3xy+23x2-3xy                =tan-12y2+2x2-2xy23y2+23x2-23xy                =tan-113=π6



Page No 4.121:

Question 15:

Let f (x) = ecos-1sin x+π/3. Then, f (8π/9) =
(a) e5π/18
(b) e13π/18
(c) e−2π/18
(d) none of these

Answer:

(b) e13π/18

Given: fx=ecos-1sinx+π3

Then,
f8π9=ecos-1sin8π9+π3=ecos-1sin11π9=ecos-1cosπ2+13π18     cosπ2+θ=sinθ=ecos-1cos13π18=e13π18

Page No 4.121:

Question 16:

tan-1111+tan-1211 is equal to
(a) 0
(b) 1/2
(c) − 1
(d) none of these

Answer:

(d) none of these

We know that tan-1x+tan-1y=tan-1x+y1-xy.
Now,
tan-1111+tan-1211=tan-1111+2111-111211=tan-1311121-2121=tan-1311119121=tan-133119=0.27

Page No 4.121:

Question 17:

If cos-1x2+cos-1y3=θ, then 9x2 − 12xy cos θ + 4y2 is equal to
(a) 36
(b) −36 sin2 θ
(c) 36 sin2 θ
(d) 36 cos2 θ

Answer:

(c) 36 sin2 θ

We know
cos-1x+cos-1y=cos-1xy-1-x21-y2

Now,
cos-1x2+cos-1y3=θcos-1x2y3-1-x241-y23=θx2y3-1-x241-y23=cosθxy-4-x29-y2=6cosθ4-x29-y2=xy-6cosθ4-x29-y2=x2y2+36cos2θ-12xycosθ               (Squaring both the sides)36-4y2-9x2+x2y2=x2y2+36cos2θ-12xycosθ36-4y2-9x2=36cos2θ-12xycosθ9x2-12xycosθ+4y2=36-36cos2θ9x2-12xycosθ+4y2=36sin2θ

Page No 4.121:

Question 18:

If tan−1 3 + tan−1 x = tan−1 8, then x =
(a) 5
(b) 1/5
(c) 5/14
(d) 14/5

Answer:

(b) 15
We know that tan-1x+tan-1y=tan-1x+y1-xy.
Now,
tan-13+tan-1x=tan-18tan-13+x1-3x=tan-183+x1-3x=83+x=8-24x3-8=-24x-x-5=-25xx=525=15

Page No 4.121:

Question 19:

The value of sin-1cos33π5 is
(a) 3π5

(b) -π10

(c) π10

(d) 7π5

Answer:

(b) -π10

sin-1cos33π5=sin-1cos6π+3π5=sin-1cos3π5=sin-1sinπ2-3π5=π2-3π5=-π10

Page No 4.121:

Question 20:

The value of cos-1cos5π3+sin-1sin5π3 is
(a) π2

(b) 5π3

(c) 10π3

(d) 0

Answer:

(d) 0

We have
cos-1cos5π3+sin-1sin5π3=cos-1cos2π-π3+sin-1sin2π-π3=cos-1cosπ3+sin-1-sinπ3=cos-1cosπ3-sin-1sinπ3=π3-π3=0

Page No 4.121:

Question 21:

sin 2 cos-1-35 is equal to
(a) 625

(b) 2425

(c) 45

(d) -2425

Answer:

(d) -2425

Let cos-1-35=x, 0xπ
Then, cos x=-35

 sinx=1-cos2x=1--352=1625=45
Now,
sin2cos-1-35=sin2x=2sinx cosx=2×45×-35=-2425

Page No 4.121:

Question 22:

If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
(a) π3

(b) π2

(c) 2π3

(d) -2π3

Answer:

(a) π3

We know
sin-1sinx=x

Now,

θ=sin-1sin-600=sin-1sin720-600=sin-1sin120=sin-1sin180-120        sinx = sinπ-x=sin-1sin60=60

Page No 4.121:

Question 23:

If 3 sin-12x1+x2-4 cos-11-x21+x2+2 tan-12x1-x2=π3 is equal to
(a) 13

(b) -13

(c) 3

(d) -34

Answer:

(a) 13

Let x=tany
Then,
3sin-12tany1+tan2y-41-tan2y1+tan2y+2tan-12tany1-tan2y=π33sin-1sin 2y-4cos-1cos 2y+2tan-1tan2y=π3                                                                        sin2y=2tany1+tan2y,cos2y=1-tan2y1+tan2y and tan2y=2tany1-tan2y3×2y-4×2y+2×2y=π36y-8y+4y=π32y=π3y=π6tan-1x=π6         tan-1x=yx=tanπ6x=13

Page No 4.121:

Question 24:

If 4 cos−1 x + sin−1 x = π, then the value of x is
(a) 32

(b) 12

(c) 32

(d) 23

Answer:

(c) 32
We know that sin-1x+cos-1x=π2.

4cos-1x+sin-1x=π4cos-1x+π2-cos-1x=π3cos-1x=π-π23cos-1x=π2cos-1x=π6x=cosπ6x=32

Page No 4.121:

Question 25:

It tan-1x+1x-1+tan-1x-1x=tan-1 (−7), then the value of x is
(a) 0
(b) −2
(c) 1
(d) 2

Answer:

(d) 2

We know that tan-1x+tan-1y=tan-1x+y1-xy.

 tan-1x+1x-1+tan-1x-1x=tan-1-7tan-1x+1x-1+x-1x1-x+1x-1×x-1x=tan-1-7tan-1x2+x+x2-2x+1xx-1x2-x-x2+1xx-1=tan-1-7tan-12x2-x+1-x+1=tan-1-7

So, we get

2x2-x+1-x+1=-72x2-x+1=7x-72x2-8x+8=0x2-4x+4=0x-22=0x=2

Page No 4.121:

Question 26:

If cos-1x>sin-1x, then

(a) 12<x1
(b) 0x<12
(c)-1x<12
(d) x > 0

Answer:

cos-1x>sin-1xcos-1x>π2-cos-1x2cos-1x>π2cos-1x>π4x>cosπ4x>12
We know that the maximum value of cosine fuction is 1.
12<x1
Hence, the correct answer is option(a).

Page No 4.121:

Question 27:

In a ∆ ABC, if C is a right angle, then
tan-1ab+c+tan-1bc+a=

(a) π3

(b) π4

(c) 5π2

(d) π6

Answer:

(b) π4

We know
 tan-1x+tan-1y=tan-1x+y1-xy

 tan-1ab+c+tan-1bc+a=tan-1ab+c+bc+a1-ab+c×bc+a                                                          =tan-1ac+a2+b2+bcb+cc+aac+c2+bcb+cc+a

=tan-1ac+c2+bcac+c2+bc     a2+b2=c2  =tan-11=tan-1tanπ4=π4

Page No 4.121:

Question 28:

The value of sin14sin-1638 is
(a) 12

(b) 13

(c) 122

(d) 133

Answer:

(c) 122

Let sin-1638=y
Then,
siny=638cosy=1-sin2y=1-6364=18
Now, we have

sin14sin-1638=sin14y=1-cosy22  cos2x=1-2sin2x=1-1+cosy22 cos2x=2cos2x-1=1-1+1822=1-9162=1-342=18=122



Page No 4.122:

Question 29:

cotπ4-2 cot-13=
(a) 7
(b) 6
(c) 5
(d) none of these

Answer:

(a) 7

Let 2cot-13=y
Then, coty2=3
 
cotπ4-2cot-13=cotπ4-y=cotπ4coty+1coty-cotπ4=coty+1coty-1     =cot2y2-12coty2+1cot2y2-12coty2-1=cot2y2+2coty2-1cot2y2-2coty2-1=9+6-19-6-1=7

Page No 4.122:

Question 30:

If tan−1 (cot θ) = 2 θ, then θ =

(a) ±π3

(b) ±π4

(c) ±π6

(d) none of these

Answer:

(c) ±π6


We have,tan-1cotθ=2θtan2θ=cotθ2tanθ1-tan2θ=1tanθ2tan2θ=1-tan2θ3tan2θ=1tan2θ=13tanθ=±13 θ=±π6

Page No 4.122:

Question 31:

If sin-12a1-a2+cos-11-a21+a2=tan-12x1-x2, where a, x0, 1, then, the value of x is

(a) 0
(b) a2
(c) a
(d) 2a1-a2

Answer:

sin-12a1-a2+cos-11-a21+a2=tan-12x1-x22tan-1a+2tan-1a=2tan-1x4tan-1a=2tan-1x2tan-1a=tan-1xtan-12a1-a2=tan-1xx=2a1-a2
Hence, the correct answer is option(d).

Page No 4.122:

Question 32:

The value of  sin2tan-10.75is equal to
(a) 0.75
(b) 1.5
(c) 0.96
(d) sin-11.5

Answer:

sin2tan-10.75=sin2tan-10.75=sinsin-12×0.751+0.752=sinsin-10.96=0.96
Hence, the correct answer is option (c).

Page No 4.122:

Question 33:

If x > 1, then 2tan-1x+sin-12x1+x2is equal to
(a) 4tan-1x
(b) 0
(c) π2
(d) π

Answer:

2tan-1x+sin-12x1+x2=2tan-1x+2tan-1x    sin-12x1+x2=2tan-1x=4tan-1x
Hence, the correct answer is option (a)

Page No 4.122:

Question 34:

The domain of cos-1x2-4 is
(a) [3, 5]
(b) [−1, 1]
(c) -5, -33, 5
(d) -5, -33, 5

Answer:

The domain of cos-1x is [−1, 1]
-1x2-41-1+4x2-4+41+43x25±3x±5x-5, -33, 5
Hence, the correct answer is option (c).

Page No 4.122:

Question 35:

The value of tancos-135+tan-114
(a) 198
(b) 819
(c) 1912
(d) 34

Answer:

tancos-135+tan-114=tantan-11-92535+tan-114=tantan-14535+tan-114=tantan-143+tan-114=tantan-143+141-13=16+31223=198
Hence, the correct answer is option (a).



Page No 4.14:

Question 1:

Find the principal values of each of the following:

(i) tan-113
(ii) tan-1-13

(iii) tan-1cosπ2

(iv) tan-12cos2π3

Answer:

(i) Let tan-113=y
Then, 
tany=13
We know that the range of the principal value branch is -π2,π2.
Thus, 
tany=13=tanπ6y=π6-π2,π2
Hence, the principal value of tan-113 is π6.

(ii) We have tan-1-13=-tan-113          tan-1-x=-tan-1x
Let tan-113=y
Then, 
tany=13
We know that the range of the principal value branch is -π2,π2.
Thus, 
tany=13=tanπ6y=π6tan-1-13=-tan-113=-y=-π6-π2,π2
Hence, the principal value of tan-1-13 is -π6.

(iii) Let tan-1cosπ2=y
Then, 
tany=cosπ2
We know that the range of the principal value branch is -π2,π2.
Thus, 
tany=cosπ2=0=tan0y=0-π2,π2
Hence, the principal value of tan-1cosπ2 is 0.

(iv)  Let tan-12cos2π3=y
Then, 
tany=2cos2π3
We know that the range of the principal value branch is -π2,π2.
Thus, 
tany=2cos2π3=2×-12=-1=tan-π4y=-π4-π2,π2
Hence, the principal value of tan-12cos2π3 is -π4.

Page No 4.14:

Question 2:

For the principal values, evaluate each of the following:

i tan-1(-1)+cos-1-12
(ii) tan-12sin4cos-132

Answer:

i tan-1-1+cos-1-12=tan-1tan-π4+cos-1cos3π4       Range of tan is -π2, π2 ;  -π4  -π2, π2 and range of cosine is 0, π ; 3π4 0, π=-π4+3π4=π2
tan-1-1+cos-1-12=π2

(ii)
tan-12sin4cos-132=tan-12sin4cos-1cosπ6=tan-12sin4×π6=tan-12sin2π3=tan-12×32=tan-13=tan-1tanπ3=π3

Page No 4.14:

Question 3:

Evaluate each of the following:

i tan-11+cos-1-12+sin-1-12
(ii) tan-1-13+tan-1-3+tan-1sin-π2

(iii) tan-1tan5π6+cos-1cos13π6

Answer:

(i)  Let sin-1-12=y
Then, 
siny=-12
We know that the range of the principal value branch is -π2,π2.
Thus, 
siny=-12=sin-π6y=-π6-π2,π2
Now,
Let cos-1-12=z
Then, 
cosz=-12
We know that the range of the principal value branch is 0,π.
Thus, 
cosz=-12=cos2π3z=2π30,π
So,
 tan-11+cos-1-12+sin-112=π4+2π3-π6  =3π4
tan-11+cos-1-12+sin-112=3π4

(ii)
tan-1-13+tan-1-3+tan-1sin-π2=tan-1-13+tan-1-3+tan-1-sinπ2=tan-1-13+tan-1-3+tan-1-1=-tan-113-tan-13-tan-11=-tan-1tanπ6-tan-1π3-tan-1π4=-π6-π3-π4=-3π4

(iii)
tan-1tan5π6+cos-1cos13π6=tan-1tanπ-5π6+cos-1cos2π+π6=tan-1-tanπ6+cos-1cosπ6=-tan-1tanπ6+cos-1cosπ6=-π6+π6=0



Page No 4.18:

Question 1:

Find the principal values of each of the following:

(i) sec-1(-2)
(ii) sec-1(2)
(iii) sec-12sin3π4
(iv) sec-12tan3π4

Answer:

(i)  Let sec-1-2=y
Then, 
secy=-2
We know that the range of the principal value branch is 0,π-π2.
Thus, 
secy=-2=sec3π4y=3π40,π, yπ2
Hence, the principal value of sec-1-2 is 3π4.

(ii) Let 
sec-12=y
Then, 
secy=2
We know that the range of the principal value branch is 0,π-π2.
Thus, 
secy=2=secπ3y=π30, π, yπ2
Hence, the principal value of sec-12 is π3.

(iii)
 Let 
sec-12sin3π4=y
Then, 
secy=2sin3π4
We know that the range of the principal value branch is 0,π-π2.
Thus, 
secy=2sin3π4=2×12=2=secπ4y=π40,π
Hence, the principal value of sec-12sin3π4 is π4.
(iv)
Let sec-12tan3π4=y
Then, 
secy=2tan3π4
We know that the range of the principal value branch is 0,π-π2.
Thus, 
secy=2tan3π4=2×-1=-2=sec2π3y=2π30,π
Hence, the principal value of sec-12tan3π4 is 2π3.

Page No 4.18:

Question 2:

For the principal values, evaluate the following:

(i) tan-13-sec-1-2

(ii) sin-1-32-2sec-12tanπ6

Answer:


(i)
tan-13-sec-1-2=tan-1tanπ3-sec-1sec2π3=π3-2π3=-π3

(ii)
sin-1-32-2sec-12tanπ6=-sin-132-2sec-12×13=-sin-132-2sec-123=-sin-1sinπ3-2sec-1secπ6=-π3-π3=-2π3

Page No 4.18:

Question 3:

Find the domain of
(i) sec-13x-1
(ii) sec-1x-tan-1x

Answer:

(ii)

Let f(x) = g(x) − h(x), where g(x)=cotx and h(x)=cot1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [0,  π/2) ⋃ [ π, 3π/2)
The domain of h(x) is -π2, π2
Therfore, the intersection of g(x) and h(x) is R − { nπ, n ⋵ Z}



Page No 4.21:

Question 1:

Find the principal values of each of the following:

(i)  cosec-1(-2)
(ii) cosec-1(-2)
(iii) cosec-123
(iv) cosec-12cos2π3

Answer:

(i)  Let cosec-1-2=y
Then, 
cosecy=-2
We know that the range of the principal value branch is -π2, π2-0.
Thus, 
cosecy=-2=cosec-π4y=-π4-π2,π2, y0
Hence, the principal value of cosec-1-2 is -π4.

(ii)
 Let 
cosec-1-2=y
Then, 
cosecy=-2
We know that the range of the principal value branch is -π2, π2-0.
Thus, 
cosecy=-2=cosec-π6y=-π6-π2,π2, y0
Hence, the principal value of cosec-1-2 is -π6.

(iii) Let cosec-123=y
Then, 
cosecy=23
We know that the range of the principal value branch is -π2, π2-0.
Thus, 
cosecy=23=cosecπ3y=π3-π2, π2, y0
Hence, the principal value of cosec-123 is π3.

(iv) 
Let 
cosec-12cos2π3=y
Then, 
cosecy=2cos2π3
We know that the range of the principal value branch is -π2, π2-0.
Thus, 
cosecy=2cos2π3=2×-12=-1=cosec-π2y=-π2-π2, π2, y0
Hence, the principal value of cosec-12cos2π3 is -π2.

Page No 4.21:

Question 2:

Find the set of values of cosec-132

Answer:

The value of cosec-132 is undefined as it is outside the range i.e., R – (–1, 1) .

Page No 4.21:

Question 3:

For the principal values, evaluate the following:

(i) sin-1-32+cosec-1-23
(ii) sec-12+2cosec-1-2
(iii) sin-1cos2cosec-1-2
(iv) cosec-12tan11π6

Answer:

(i)
sin-1-32+cosec-1-23=-sin-132+cosec-1-23=-sin-1sinπ3+cosec-1cosec-π3=-π3-π3=-2π3
(ii)
sec-12+2cosec-1-2=sec-1secπ4+2cosec-1cosec-π4=π4-2×π4=π4-π2=-π4
(iii)

sin-1cos2cosec-1-2=sin-1cos2cosec-1cosec-π6=sin-1cos-π3=sin-1cosπ3=sin-112=sin-1sinπ6=π6
(iv)
cosec-12tan11π6=cosec-12×-13=cosec-1-23=cosec-1cosec-π3=-π3



Page No 4.24:

Question 1:

Find the principal values of each of the following:

(i) cot-1(-3)
(ii) cot-13
(iii) cot-1-13
(iv) cot-1tan3π4

Answer:

(i)  Let cot-1-3=y
Then, 
coty=-3
We know that the range of the principal value branch is 0,π.
Thus, 
coty=-3=cot5π6y=5π60,π
Hence, the principal value of cot-1-3 is 5π6.

(ii) Let 
cot-13=y
Then, 
coty=3
We know that the range of the principal value branch is 0,π.
Thus, 
coty=3=cotπ6y=π60,π
Hence, the principal value of cot-13 is π6.
(iii) Let cot-1-13=y
Then, 
coty=-13
We know that the range of the principal value branch is 0,π.
Thus, 
coty=-13=cot2π3y=2π30,π
Hence, the principal value of cot-1-13 is 2π3.
(iv) 
Let cot-1tan3π4=y
Then, 
coty=tan3π4
We know that the range of the principal value branch is 0,π.
Thus, 
coty=tan3π4=-1=cot3π4y=3π40,π
Hence, the principal value of cot-1tan3π4 is 3π4.

Page No 4.24:

Question 2:

Find the domain of fx=cotx+cot-1x

Answer:

Let f(x) = g(x) + h(x), where gx=cotx and hx=cot-1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is R − { nπ, n ⋵ Z}
The domain of h(x) is (0, π )
Therfore, the intersection of g(x) and h(x) is R − { nπ, n ⋵ Z}

Page No 4.24:

Question 3:

Evaluate each of the following:

(i) cot-113-cosec-1-2+sec-123

(ii) cot-12cossin-132

(iii) cosec-1-23+2cot-1-1

(iv) tan-1-13+cot-113+tan-1sin-π2

Answer:

(i)
cot-113-cosec-1-2+sec-123=cot-1cotπ3-cosec-1cosec-π6+sec-1secπ6=π3+π6+π6=2π3

(ii)
cot-12cossin-132=cot-12cossin-1sinπ3=cot-12cosπ3=cot-12×12=cot-11=cot-1tanπ4=π4

(iii)
cosec-1-23+2cot-1-1=cosec-1cosec-π3+2cot-1cot3π4=-π3+2×3π4=-π3+3π2=7π6

(iv)
tan-1-13+cot-113+tan-1sin-π2=tan-1tan-π6+cot-1cotπ3+tan-1-1=tan-1tan-π6+cot-1cotπ3+tan-1tan-π4=-π6+π3-π4=-π12



Page No 4.42:

Question 1:

(i) sin-1sinπ6
(ii) sin-1sin7π6
(iii)  sin-1sin5π6
(iv) sin-1sin13π7
(v) sin-1sin17π8
(vi) sin-1sin-17π8
(vii) sin-1sin3
(viii) sin-1sin4
(ix) sin-1sin12
(x) sin-1sin2

Answer:

We know

sinsin-1θ=θ if -π2θπ2
(i) We have

sin-1sinπ6=π6

(ii) We have

sin-1sin7π6=sin-1sinπ+π6=sin-1sin-π6=-π6

(iii) We have

sin-1sin5π6=sin-1sinπ-π6=sin-1sinπ6=π6

(iv) We have

sin-1sin13π7=sin-1sin2π-π7=sin-1sin-π7=-π7

(v) We have

sin-1sin17π8=sin-1sin2π+π8=sin-1sinπ8=π8

(vi) We have

sin-1sin-17π8=sin-1-sin17π8=sin-1-sin2π+π8=sin-1-sinπ8=sin-1sin-π8=-π8

(vii) We have

sin-1sin3=sin-1sinπ-3=π-3

(viii)We have

sin-1sin4=sin-1sinπ-4=π-4

(ix) We have

sin-1sin12=sin-1sin-π+12=12-π

(x) )We have

sin-1sin2=sin-1sinπ-2=π-2

Page No 4.42:

Question 2:

Evaluate each of the following:
(i) 
cos-1cos-π4
(ii) cos-1cos5π4
(iii)  cos-1cos4π3
(iv) cos-1cos13π6 
(v) cos-1cos3

(vi) cos-1cos4

(vii) cos-1cos5

(viii) cos-1cos12

Answer:

We know

cos-1cosθ=θ if 0θπ

(i)  We have

cos-1cos-π4=cos-1cosπ4=π4

(ii) We have

cos-1cos5π4=cos-1cos2π-3π4=cos-1cos3π4=3π4

(iii) We have

cos-1cos4π3=cos-1cos2π-2π3=cos-1cos2π3=2π3

(iv) We have

cos-1cos13π6=cos-1cos2π+π6=cos-1cosπ6=π6

(v) We have

cos-1cos3=3

(vi)We have

cos-1cos4=cos-1cos2π-4=2π-4

(vii) We have

cos-1cos5=cos-1cos2π-5=2π-5

(viii) We have

cos-1cos12=cos-1cos4π-12=4π-12
 

Page No 4.42:

Question 3:

Evaluate each of the following:

(i) tan-1tanπ3
(ii) tan-1tan6π7
(iii) tan-1tan7π6
(iv) tan-1tan9π4
(v) tan-1tan1
(v) tan-1tan2
(v) tan-1tan4
(v) tan-1tan12

Answer:

We know that

tan1tanθ=θ,      -π2<θ<π2

(i) We have 
tan-1tanπ3=π3
(ii) We have
tan-1tan6π7=tan-1tanπ-π7=tan-1tan-π7=-π7
(iii) We have

tan-1tan7π6=tan-1tanπ+π6=tan-1tanπ6=π6
(iv) We have

tan-1tan9π4=tan-1tan2π+π4=tan-1tanπ4=π4
(v) We have
tan-1tan1=1
(vi) We have

tan-1tan2=tan-1tan-π+2=2-π

(vii) We have

tan-1tan4=tan-1tan-π+4=4-π

(viii) We have

tan-1tan12=tan-1tan-4π+12=12-4π

Page No 4.42:

Question 4:

Evaluate each of the following:

(i) sec-1secπ3
(ii) sec-1sec2π3
(iii) sec-1sec5π4
(iv) sec-1sec7π3

(v) sec-1sec9π5
(vi)  sec-1sec-7π3
(vii) sec-1sec13π4
(viii) sec-1sec25π6

Answer:

We know that

sec1secθ=θ,      [0, π/2) ⋃ ( π/2,  π]

(i) We have 
sec-1secπ3=π3
(ii) We have
sec-1sec2π3=2π3
(iii) We have

sec-1sec5π4=sec-1sec2π-3π4=sec-1sec3π4=3π4
(iv)We have

sec-1sec7π3=sec-1sec2π+π3=sec-1secπ3=π3

(v)We have

sec-1sec9π5=sec-1sec2π-π5=sec-1secπ5=π5
(vi) We have

sec-1sec-7π3=sec-1sec7π3=sec-1sec2π+π3=sec-1secπ3=π3
(vii)We have

sec-1sec13π4=sec-1sec4π-3π4=sec-1sec3π4=3π4
(viii)We have

sec-1sec25π6=sec-1sec4π+π6=sec-1secπ6=π6

Page No 4.42:

Question 5:

Evaluate each of the following:

(i) cosec-1cosecπ4
(ii) cosec-1cosec3π4
(iii) cosec-1cosec6π5
(iv) cosec-1cosec11π6
(v) cosec-1cosec13π6
(vi) cosec-1cosec-9π4

Answer:

We know that

cosec-1cosecθ=θ,      [−π/2, 0) ⋃ ( 0, π/2]

(i) cosec-1cosecπ4=π4
(ii) 
cosec-1cosec3π4=cosec-1cosecπ-π4=cosec-1cosecπ4=π4
(iii)
cosec-1cosec6π5=cosec-1cosecπ+π5=cosec-1cosec-π5=-π5
(iv)
cosec-1cosec11π6=cosec-1cosec2π-π6=cosec-1cosec-π6=-π6
(v)
cosec-1cosec13π6=cosec-1cosec2π+π6=cosec-1cosecπ6=π6
(vi)
cosec-1cosec-9π4=cosec-1-cosec2π+π4=cosec-1-cosecπ4=cosec-1cosec-π4=-π4



Page No 4.43:

Question 6:

Evaluate each of the following:

(i) cot-1cotπ3
(ii) cot-1cot4π3
(iii) cot-1cot9π4
(iv) cot-1cot19π6
(v) cot-1cot-8π3
(vi) cot-1cot21π4

Answer:

We know that

cot-1cotθ=θ,      ( 0, π)

(i) We have
cot-1cotπ3=π3
(ii) We have
cot-1cot4π3=cot-1cotπ+π3=cot-1cotπ3=π3
(iii) We have
cot-1cot9π4=cot-1cot2π+π4=cot-1cotπ4=π4
(iv) We have
cot-1cot19π6=cot-1cotπ+π6=cot-1cotπ6=π6
(v) We have
cot-1cot-8π3=cot-1-cot8π3=cot-1-cot3π-π3=cot-1cotπ3=π3
(vi) We have
cot-1cot21π4=cot-1cot5π+π4=cot-1cotπ4=π4
 

Page No 4.43:

Question 7:

Write each of the following in the simplest form:

(i)  
cot-1ax2-a2,  x >a
(ii) tan-1x+1+x2, xR

(iii) tan-11+x2-x, xR

(iv) tan-11+x2-1x, x0

(v) tan-11+x2+1x, x0

(vi) tan-1a-xa+x,-a<x<a

(vii) tan-1xa+a2-x2,-a<x<a

(viii) sin-1x+1-x22,-1<x<1

(ix) sin-11+x+1-x2, 0<x<1

(x) sin2 tan-11-x1+x

Answer:

(i) Let x=asecθ
Now,

cot-1ax2-a2=cot-1aa2sec2θ-a2=cot-1aatan2θ=cot-1cotθ=θ  =sec-1xa  

(ii) Let x=cotθ
Now,
tan-1x+1+x2=tan-1cotθ+1+cot2θ=tan-1cotθ+cosecθ   =tan-1cosθ+1sinθ=tan-12cos2θ22sinθ2cosθ2=tan-1cotθ2=tan-1tanπ2-θ2 =π2-θ2=π2-cot-1x2



(iii) Let x=cotθ
Now,

tan-11+x2-x=tan-11+cot2θ-cotθ=tan-1cosecθ-cotθ=tan-11-cosθsinθ=tan-12sin2θ22sinθ2cosθ2=tan-1tanθ2=θ2=cot-1x2

(iv) Let x=tanθ
Now,

tan-11+x2-1x=tan-11+tan2θ-1tanθ=tan-1sec2θ-1tanθ  =tan-1secθ-1tanθ=tan-11-cosθsinθ=tan-12sin2θ22sinθ2cosθ2=tan-1tanθ2=θ2=tan-1x2

(v) Let x=tanθ
Now,

tan-11+x2+1x=tan-11+tan2θ+1tanθ=tan-1sec2θ+1tanθ     =tan-1secθ+1tanθ=tan-1cosθ+1sinθ=tan-12cos2θ22sinθ2cosθ2=tan-1cotθ2=tan-1tanπ2-θ2 =π2-θ2=π2-tan-1x2


(vi) Let x=acosθ
Now,

tan-1a-xa+x=tan-1a-acosθa+acosθ=tan-11-cosθ1+cosθ=tan-12sin2θ22cos2θ2=tan-1tanθ2=θ2 =12cos-1xa
  tan-1a-xa+x=cos-1xa2

(vii) Let x=asinθ
Now,

tan-1xa+a2-x2=tan-1asinθa+a2-a2cos2θ=tan-1asinθa+acos2θ=tan-1sinθ1+cosθ=tan-12sinθ2cosθ22cos2θ2=tan-1tanθ2   =θ2=12sin-1xa


(viii) Let x=sinθ
Now,

sin-1x+1-x22=sin-1sinθ+1-sin2θ2=sin-1sinθ+cosθ2=sin-112sinθ+12cosθ=sin-1cosπ4sinθ+sinπ4cosθ=sin-1sinθ+π4   =θ+π4=π4+sin-1x
  sin-1x+1-x22=cos-1x+π4

(ix) Let x=cosθ
Now,
sin-11+x+1-x2=sin-11+cosθ+1-cosθ2=sin-12cos2θ2+2sin2θ22=sin-1cosθ2+sinθ22=sin-112sinθ2+12cosθ2=sin-1sinθ2+π4    =θ2+π4=cos-1x2+π4
  sin-11+x+1-x2=cos-1x2+π4

(x) Let x=cosθ
Now,

sin2tan-11-x1+x=sin2tan-11-cosθ1+cosθ=sin2tan-12sin2θ22cos2θ2=sin2tan-1tanθ2=sinθ=sincos-1x   =sinsin-11-x2=1-x2



Page No 4.54:

Question 1:

Evaluate each of the following:

(i) sinsin-1725
(ii) sincos-1513
(iii) sintan-1247
(iv) sinsec-1178
(v) coseccos-135
(vi) secsin-11213

(vii) tancos-1817
(viii) cotcos-135
(ix) costan-1247

Answer:

(i) sinsin-1725=725
(ii)

sincos-1513=sinsin-11-5132        cos-1x=sin-11-x2=sinsin-11-25169=sinsin-1144169=sinsin-11213=1213
(iii)
sintan-1247=sinsin-12471+2472      tan-1x=x1+x2=sinsin-12471+57649=sinsin-124762549=sinsin-1247257=2425
(iv)
sinsec-1178=sincos-1817=sinsin-11-8172        cos-1x=sin-11-x2=sinsin-11-64289=sinsin-1225289=sinsin-11517=1517
(v) 
coseccos-135=cosecsin-11-352        cos-1x=sin-11-x2=cosecsin-11-925=cosecsin-11625=cosecsin-145=coseccosec-154=54
(vi)
secsin-11213=seccos-11-12132        sin-1x=cos-11-x2=seccos-11-144169=seccos-125169=seccos-1513=secsec-1135=135

(vii)
 
tancos-1817=tantan-11-8172817    cos-1x=tan-11-x2x                        =tantan-11517817                        =158
(viii)
cotcos-135=cottan-11-35235    cos-1x=tan-11-x2x=cottan-14535=cotcot-134=34
(ix) 
costan-1247=coscos-111+2472       tan-1x=cos-111+x2=coscos-111+57649=coscos-11257=coscos-1725=725

Page No 4.54:

Question 2:

Prove the following results

(i) tancos-145+tan-123=176
(ii) cossin-135+cot-132=6513

(iii) tansin-1513+cos-135=6316

(iv) sincos-135+sin-1513=6365

Answer:

(i)
LHS=tancos-145+tan-123=tantan-11-45245+tan-123    cos-1x=tan-11-x2x                                           =tantan-134+tan-123                                           =tantan-134+231-34×23          tan-1x+tan-1y=tan-1x+y1-xy                                           =tantan-11712612                                           =tantan-1176                                           =176=RHS                                          

(ii)
LHS=cossin-135+cot-132=cossin-135+tan-123=coscos-11-352+cos-111+232=coscos-145+cos-1313=coscos-145×313-1-4521-3132=coscos-112513-6513=coscos-16513=6513=RHS


(iii)
The question is wrong as we can't have arc sin greater than 1

(iv) 
LHS=sincos-135+sin-1513=sinsin-11-352+sin-1513=sinsin-145+sin-1513=sinsin-145×1-5132+513×1-452=sinsin-14865+1565=sinsin-16365=6365=RHS

Page No 4.54:

Question 3:

Solve: cossin-1x=16

Answer:

cossin-1x=16coscos-11-x2=161-x2=161-x2=1361-136=x2x2=3536x=±356

Page No 4.54:

Question 4:

Solve: cos2sin-1-x=0

Answer:

Given,
cos2sin-1-x=0cos-2sin-1x=0                                             sin-1-θ=-sin-1θcos2sin-1x=0                                                cos-θ=cosθ
We know, -π2sin-1θπ2
Therefore, 2sin-1x=±π2
sin-1x=±π4x=±12



Page No 4.58:

Question 1:

Evaluate:
(i) cossin-1-725
(ii) seccot-1-512
(iii) cotsec-1-135

Answer:

(i)
cossin-1-725=cos-sin-1725=cossin-1725=coscos-11-7252=coscos-12425=2425
(ii)
seccot-1-512=secπ-cot-1512=-seccot-1512=-seccos-111+1252=-seccos-1513=-secsec-1135=-135
(iii)
cotsec-1-135=cotsec-1π-135=-cotsec-1135=-cottan-11-5133513=-cottan-1125=-cotcot-1512=-512

Page No 4.58:

Question 2:

Evaluate:
(i) tancos-1-725
(ii) coseccot-1-125
(iii) costan-134

Answer:

(i) 
tancos-1-725=tancos-1π-725=-tancos-1725=-tantan-11-7252725=-tantan247=-247
(ii) 
coseccot-1-125=coseccot-1π-125=coseccot-1125=cosecsin-15121+5122=cosecsin-1513=coseccosec-1135=135
(iii)  We have

costan-134=cos12cos-11-3421+342   2tan-1x=cos-11-x21+x2=cos12cos-1725
Let y=cos-1725cosy=725
Now,

cos12cos-1725=cos12y=cosy+12    cos2x=2cos2x-1=725+12=3250=45
costan-134=45



Page No 4.59:

Question 3:

Evaluate: sincos-1-35+cot-1-512

Answer:

sincos-1-35+cot-1-512=sinπ-cos-135+π-cot-1512=sin2π-cos-135+cot-1512=-sincos-135+cot-1512=-sinsin-11-352+sin-11251+1252=-sinsin-145+sin-11213=-sinsin-145×1-12132+1213×1-452=-sinsin-12065+3665=-sinsin-15665=-5665



Page No 4.6:

Question 1:

Find the principal value of each of the following:

(i) sin-1-32
(ii) sin-1cos2π3
(iii) sin-13-122
(iv) sin-13+122
(v) sin-1cos3π4
(vi) sin-1tan5π4

Answer:

(i) sin-1-32=sin-1sin-π3=-π3
(ii) sin-1cos2π3=sin-1-12=sin-1sin-π6=-π6
(iii) sin-13-122=sin-1sinπ12=π12
(iv) sin-13+122=sin-1sin5π12=5π12
(v) sin-1cos3π4=sin-1-22=sin-1sin-π4=-π4
(vi) sin-1tan5π4==sin-11=sin-1sinπ2=π2



Page No 4.66:

Question 1:

Evaluate:

(i) cotsin-134+sec-143
(ii) sintan-1x+tan-11x for x<0
(iii) sintan-1x+tan-11x for x>0
(iv) cottan-1a+cot-1a
(v) cossec-1x+cosec-1x, x1 

Answer:

(i)
cotsin-134+sec-143=cotsin-134+cos-134       sec-1x=cos-11x   =cotπ2         sin-1x+cos-1x=π2=0

(ii)
sintan-1x+tan-11x=sintan-1-x+tan-1-1x     x<0=sin-tan-1x-tan-11x=sin-tan-1x+tan-11x=sin-tan-1x+cot-1x        tan-11x=cot-1x=-sintan-1x+cot-1x        =-sinπ2         tan-1x+cot-1x=π2=-1

(iii)
sintan-1x+tan-11x=sintan-1x+cot-1x       tan-1x=cot-11x   =sinπ2         tan-1x+cot-1x=π2=1

(iv)
cottan-1a+cot-1a    =cotπ2         tan-1x+cot-1x=π2=0

(v)
cossec-1x+cosec-1x    =cosπ2         sec-1x+cosec-1x=π2=0

Page No 4.66:

Question 2:

If cos-1x+cos-1y=π4, find the value of sin-1x+sin-1y

Answer:

cos-1x+cos-1y=π4π2-sin-1x+π2-sin-1y=π4        cos-1x=π2-sin-1xπ-sin-1x+sin-1y=π4sin-1x+sin-1y=3π4

Page No 4.66:

Question 3:

If sin-1x+sin-1y=π3 and cos-1x-cos-1y=π6, find the values of x and y.

Answer:

cos-1x-cos-1y=π6π2-sin-1x-π2+sin-1y=π6        cos-1x=π2-sin-1x-sin-1x-sin-1y=π6sin-1x-sin-1y=-π6
Solving sin-1x+sin-1y=π3 and sin-1x-sin-1y=-π6, we will get
2sin-1x=π6sin-1x=π12x=sinπ12=3-122andsin-1y=π3-sin-1xsin-1y=π3-π12sin-1y=π4y=sinπ4=12

Page No 4.66:

Question 4:

If cotcos-135+sin-1x=0, find the values of x.

Answer:

cotcos-135+sin-1x=0cos-135+sin-1x=cot0 cos-135+sin-1x=π2cos-135=π2-sin-1xcos-135=cos-1x       cos-1x=π2-sin-1xx=35

Page No 4.66:

Question 5:

If sin-1x2+cos-1x2=17π236, find x

Answer:

sin-1x2+cos-1x2=17π236sin-1x2+π2-sin-1x2=17π236Let sin-1x=yy2+π2-y2=17π236
y2+π24+y2-2×π2×y=17π2362y2-πy=2π2918y2-9πy-2π2=018y2-12πy+3πy-2π2=06y3y-2π+π3y+2π=03y-2π6y+π=0y=-π6     Neglecting y=23π as it is not satisfying the questionx=siny=sin-π6=-12

Page No 4.66:

Question 6:

sinsin-115+cos-1x=1

Answer:

sinsin-115+cos-1x=1sin-115+cos-1x=sin-11    sin-115+cos-1x=π2sin-115=π2-cos-1xsin-115=sin-1x                   sin-1x=π2-cos-1xx=15

Page No 4.66:

Question 7:

sin-1x=π6+cos-1x

Answer:

sin-1x=π6+cos-1xsin-1x=π6+π2-sin-1x      cos-1x=π2-sin-1x2sin-1x=2π3sin-1x=π3sin-1x=π3x=sinπ3=32

Page No 4.66:

Question 8:

4sin-1x=π-cos-1x

Answer:

4sin-1x=π-cos-1x4sin-1x=π-π2-sin-1x      cos-1x=π2-sin-1x4sin-1x=π2+sin-1x 3sin-1x=π2sin-1x=π6x=sinπ6=12

Page No 4.66:

Question 9:

tan-1x+2cot-1x=2π3

Answer:

tan-1x+2cot-1x=2π3tan-1x+2π2-tan-1x=2π3      cot-1x=π2-tan-1xtan-1x+π-2tan-1x=2π3 tan-1x=π3tan-1x=π3x=tanπ3=3

Page No 4.66:

Question 10:

5tan-1x+3cot-1x=2π

Answer:

5tan-1x+3cot-1x=2π5tan-1x+3π2-tan-1x=2π      cot-1x=π2-tan-1x5tan-1x+3π2-3tan-1x=2π 2tan-1x=π2tan-1x=π4x=tanπ4=1



Page No 4.7:

Question 2:

(i) sin-112-2sin-112
(ii) sin-1cossin-132

Answer:

(i)

sin-112-2sin-112=sin-112-sin-12×121-122=sin-112-sin-12×12=sin-112-sin-11=sin-1sinπ6-sin-1sinπ2=π6-π2=-π3
(ii)
sin-1cossin-132=sin-1cossin-1sinπ3=sin-1cosπ3=sin-112=sin-1sinπ6=π6

Page No 4.7:

Question 3:

Find the domain of each of the following functions:

i fx=sin-1x2ii fx=sin-1x+sinxiii fxsin-1x2-1iv fx=sin-1x+sin-12x

Answer:

(i)
To the domain of sin-1y which is [−1, 1]
x20, 1 as x2 can not be negative
x-1, 1
Hence, the domain is [−1, 1]

(ii)
Let f(x) = g(x) + h(x), where g(x)=cotx and h(x)=cot1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [−1, 1]
The domain of h(x) is (−∞, ∞)
Therfore, the intersection of g(x) and h(x) is [−1, 1]
Hence, the domain is [−1, 1].

(iii)
To the domain of sin-1y which is [−1, 1]
x2-10, 1 as square root can not be negative
x21, 2x-2, -11, 2
Hence, the domain is -2, -11, 2

(iv)
Let f(x) = g(x) + h(x), where g(x)=cotx and h(x)=cot1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [−1, 1]
The domain of h(x) is -12, 12
Therfore, the intersection of g(x) and h(x) is -12, 12
Hence, the domain is -12, 12

 

Page No 4.7:

Question 4:

If sin-1x+sin-1y+sin-1z+sin-1t=2π, then find the value of x2 + y2 + z2 + t2

Answer:

We know that the maximum value of sin-1x, sin-1y, sin-1z and sin-1t isπ2
Now,
LHS=sin-1x+sin-1y+sin-1z+sin-1t        =π2+π2+π2+π2        =2π=RHS
Now,
sin-1x=π2, sin-1y=π2, sin-1z=π2 and sin-1t=π2x=sinπ2, y=sinπ2, z=sinπ2 and t=sinπ2x=1, y=1, z=1 and t=1x2+y2+z2+t2=1+1+1+1=4

Page No 4.7:

Question 5:

If sin-1x2+sin-1y2+sin-1z2=34π2, find the value of x2 + y2 + z2

Answer:

We know that the maximum value of sin-1x, sin-1y, sin-1z isπ2 and minimum value of sin-1x, sin-1y, sin-1z is-π2
Now,
For maximum value
LHS=sin-1x2+sin-1y2+sin-1z2        =π22+π22+π22        =34π2=RHS
and For minimum value
LHS=sin-1x2+sin-1y2+sin-1z2        =-π22+-π22+-π22        =34π2=RHS
Now, For maximum value
sin-1x=π2, sin-1y=π2, sin-1z=π2x=sinπ2, y=sinπ2, z=sinπ2x=1, y=1, z=1x2+y2+z2=1+1+1=3
and for minimum value
sin-1x=-π2, sin-1y=-π2, sin-1z=-π2x=sin-π2, y=sin-π2, z=sin-π2x=-1, y=-1, z=-1x2+y2+z2=1+1+1=3



Page No 4.82:

Question 1:

Prove the following results:

(i) 
tan-117+tan-1113=tan-129
(ii) sin-11213+cos-145+tan-16316=π
(iii) tan-114+tan-129=sin-115

Answer:

 i LHS=tan-117+tan-1 113          =tan-117+1131-17×113          tan-1x+tan-1y=tan-1x+y1-xy          =tan-120919091          =tan-129=RHS                             

ii LHS=sin-11213+cos-1 45+tan-16316             =tan-112131-144169+tan-11-162545+tan-16316    sin-1x=tan-1x1-x2 and cos-1x=tan-11-x2x              =tan-11213513+tan-13545+tan-16316             =tan-1125+tan-134+tan-16316             =π+tan-1125+341-125×34 +tan-16316         tan-1x+tan-1y=π+tan-1x+y1-xy             =π+tan-16320-1620+tan-16316             =π+tan-1-6316+tan-16316             =π-tan-16316+tan-16316             =π=RHS

(iii)
LHS=tan-114+tan-1 29          =tan-114+291-14×29          tan-1x+tan-1y=tan-1x+y1-xy          =tan-117363436          =tan-112          =sin-1121+122          =sin-115=RHS                             
 
 

Page No 4.82:

Question 2:

Find the value of tan-1xy-tan-1x-yx+y

Answer:

We know 
tan-1x-tan-1y=tan-1x-y1+xy,xy>-1
Now,
tan-1xy-tan-1x-yx+y
=tan-1xy-x-yx+y1+xyx-yx+y=tan-1x2+xy-xy+y2yx+yx2+y2+xy-xyyx+y=tan-11=tan-1tanπ4=π4
 tan-1xy-tan-1x-yx+y=π4

Page No 4.82:

Question 3:

Solve the following equations for x:
(i) tan−12x + tan−13x = + 3π4
(ii) tan−1(x + 1) + tan−1(x − 1) = tan−1831
(iii) tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
(iv) 
 tan−11-x1+x-12tan−1x = 0, where x > 0
(v) cot
−1x − cot−1(x + 2) = π12> 0
(vi)  tan
−1(x + 2) + tan−1(x − 2) = tan−1879x > 0
(vii)  
tan-1x2+tan-1x3=π4, 0<x<6
(viii) tan-1x-2x-4+tan-1x+2x+4=π4
(ix) tan-12+x+tan-12-x=tan-123, where x<-3 or, x>3
(x) tan-1x-2x-1+tan-1x+2x+1=π4

Answer:

(i) We know
tan-1x+tan-1y=tan-1x+y1-xy

 tan-12x+tan-13x=nπ+3π4tan-12x+3x1-2x×3x=nπ+3π45x1-6x2=tannπ+3π45x1-6x2=-15x=-1+6x26x2-5x-1=06x+1x-1=0x=-16  As x=1 is not satisfying the equation


(ii)  We know
tan-1x+tan-1y=tan-1x+y1-xy

 tan-1x+1+tan-1x-1=tan-1831tan-1x+1+x-11-x+1×x-1=tan-18312x1-x2+1=8312x2-x2=83131x=8-4x24x2+31x-8=04x2+32x-x-8=04x-1x+8=0x=14     As x=-8 is not satisfying the equation
(iii) We know
tan-1x+tan-1y=tan-1x+y1-xy and tan-1x-tan-1y=tan-1x-y1+xy

 tan-1x+1+tan-1x-1+tan-1x=tan-13xtan-1x+1+x-11-x+1×x+1=tan-13x-tan-1xtan-12x2-x2=tan-13x-x1+3x22x2-x2=2x1+3x22-x2=1+3x24x2-1=0x2=14x=±12 

(iv)
 tan-11-x1+x-12tan-1x=0tan-11-x1+x=12tan-1xtan-11- tan-1x=12tan-1x     tan-11- tan-1x=tan-11-x1+xtan-11=32tan-1xπ4=32tan-1xπ6=tan-1xx=13

(v)
 cot-1x-cot-1x+2=π12tan-11x+cot-11x+2=π12   cot-1x=tan-11xtan-11x-1x+21+1xx+2=π12      tan-12xx+2x2+2x+1xx+2=π12tan-12x2+2x+1=π122x2+2x+1=tanπ12   2x2+2x+1=tanπ3-π4        2x2+2x+1=tanπ3-tanπ41+tanπ3×tanπ42x2+2x+1=3-13+12x2+2x+1=3-13+1×3+13+12x2+2x+1=23+121x+12=13+12x+1=3+1x=3

(vi) We know
tan-1x+tan-1y=tan-1x+y1-xy

 tan-1x+2+tan-1x-2=tan-1879tan-1x+2+x-21-x+2×x-2=tan-18792x1-x2+4=879x5-x2=47979x=20-4x24x2+79x-20=04x2+80x-x-20=04x-1x+20=0x=14 or- 20 x=14    x>0

(vii)  We know
tan-1x+tan-1y=tan-1x+y1-xy

 tan-1x2+tan-1x3=π4tan-1x2+x31-x2×x3=π4tan-15x66-x26=π45x6-x2=tanπ45x6-x2=15x=6-x2x2+5x-6=0x-1x+6=0x=1      0<x<6

(viii)
We know



 tan-1x-2x-4+tan-1x+2x+4=π4tan-1x-2x-4+x+2x+41-x-2x-4×x+2x+4=π4tan-1x2+2x-8+x2-2x-8x-4x+4x2-16-x2+4x-4x+4=π42x2-16-12=tanπ42x2-16-12=12x2-16=-122x2=4x2=2x=±2    tan-1x+tan-1y=tan-1x+y1-xy

(ix) 
We know
tan-1x+tan-1y=tan-1x+y1-xy

 tan-1x+2+tan-1x-2=tan-123tan-12+x+2-x1-2+x×2-x=tan-12341-4+x2=23-6+2x2=122x2=18x2=9 x=±3  

(x)
tan-1x-2x-1+tan-1x+2x+1=π4tan-1x-2x-1+x+2x+11-x-2x-1x+2x+1=π4                       tan-1x+tan-1y=tan-1x+y1-xyx-2x+1+x-1x+2x-1x+1x-1x+1-x-2x+2x-1x+1=tanπ4x-2x+1+x-1x+2x-1x+1-x-2x+2=1x2-x-2+x2+x-2x2-1-x2-4=1

2x2-43=12x2-4=32x2=7x2=72x=±72

Page No 4.82:

Question 4:

Sum the following series:
tan-113+tan-129+tan-1433+...+tan-12n-11+22n-1

Answer:

     tan-113+tan-129+tan-1433+...+tan-12n-11+22n-1tan-12-11+2×1+tan-14-21+4×2+tan-18-41+8×4+...+tan-12n-2n-11+2n.2n-1tan-12-tan-11+tan-14-tan-12+tan-18-tan-14+...+tan-12n-1-tan-12n-2+tan-12n-tan-12n-1tan-12n-tan-11tan-12n-π4



Page No 4.89:

Question 1:

Evaluate: cossin-135+sin-1513

Answer:

cossin-135+sin-1513=cossin-1351-5132+5131-352         sin-1x+sin-1y=sin-1x1-y2+y1-x2                                   =cossin-135×1213+513×45                                   =cossin-13665+413                                   =cossin-15665                                   =coscos-11-56652    sin-1x=cos-11-x2                                   =coscos-13365                                   =3365

Page No 4.89:

Question 2:

(i) sin-16365=sin-1513+cos-135
(ii) sin-1513+cos-135=tan-16316
(iii) 9π8-94sin-113=94sin-1223

Answer:

(i) 
RHS    sin-1513+cos-135= sin-1513+sin-145                cos-1x=sin-11-x2=sin-15131-452+451-5132=sin-1513×35+45×1213=sin-11565+4865=sin-16365=LHS


(ii)

LHS=sin-1513+cos-135      =sin-1513+cos-135      =sin-1513+sin-11-352    sin-1x=cos-11-x2      =sin-1513+sin-145      =sin-15131-452+451-5132                 sin-1x+sin-1y=sin-1x1-y2+y1-x2      =sin-1513×35+45×1213      =sin-1313+4865      =sin-16365     =tan-163651-63652   sin-1x=tan-1x1-x2     =tan-163651665     =tan-16316=RHS

(iii)

9π8-94sin-113=94sin-1223LHS=9π8-94sin-113        =94π2-sin-113        =94cos-113        =94sin-11-19        =94sin-1223=RHS

Page No 4.89:

Question 3:

Solve the following:

(i)  sin−1x + sin−12x = π3
(ii)  cos-1x+sin-1x2=π6
 

Answer:

(i) We know
sin-1x+sin-1y=sin-1x1-y2+y1-x2

 sin-1x+sin-12x=π3sin-1x+sin-12x=sin-132sin-1x-sin-132=-sin-12xsin-1x1-34+321-x2=-sin-12xsin-1x2+321-x2=sin-1-2xx2+321-x2=-2xx+31-x2=-4x5x=-31-x2Squaring both the sides,25x2=3-3x228x2=3x=±1237
(ii) 

cos-1x+sin-1x2=π6cos-1x+sin-1x2=sin-112cos-1x=sin-112-sin-1x2cos-1x=sin-1121-x24-x21-14    sin-1x-sin-1y=sin-1x1-y-y1-x2cos-1x=sin-13x4-3x4sin-11-x2=sin-13x4-3x41-x2=0Squaring both the sides,1-x2=0 x=±1     As x=-1 is not satisfying the equation



Page No 4.92:

Question 1:

If cos-1x2+cos-1y3=α, then prove that 9x2 − 12xy cos α + 4y2 = 36 sin2 α.

Answer:

We know
cos-1x+cos-1y=cos-1xy-1-x21-y2
Now,

cos-1x2+cos-1y3=αcos-1x2y3-1-x241-y23=αx2y3-1-x241-y23=cosαxy-4-x29-y2=6cosα4-x29-y2=xy-6cosα4-x29-y2=x2y2+36cos2α-12xycosα      Squaring both sides36-4y2-9x2+x2y2=x2y2+36cos2α-12xycosα36-4y2-9x2=36cos2α-12xycosα9x2-12xycosα+4y2=36-36cos2α9x2-12xycosα+4y2=36sin2α

Page No 4.92:

Question 2:

Solve the equation cos-1ax-cos-1bx=cos-11b-cos-11a

Answer:

cos-1ax-cos-1bx=cos-11b-cos-11acos-1ax+cos-11a=cos-11b+cos-1bxcos-1ax×1a-1-ax21-1a2=cos-1bx×1b-1-bx21-1b2     cos-1x+cos-1y=cos-1xy-1-x21-y2cos-11x-1-a2x2×1-1a2=cos-11x-1-b2x2×1-1b2 1x-1-a2x2×1-1a2=1x-1-b2x2×1-1b21-a2x21-1a2=1-b2x21-1b21-1a2-a2x2+1x2=1-1b2-b2x2+1x2a2-b2x2=1b2-1a2a2-b2x2=a2-b2a2b2x2=a2b2x=ab
 

Page No 4.92:

Question 3:

Solve  cos-13x+cos-1x=π2

Answer:

cos-13x+cos-1x=π2cos-13x×x-1-3x21-x2=π2       cos-1x+cos-1y=cos-1xy-1-x21-y2cos-13x2-1-3x21-x2=π23x2-1-3x21-x2=cosπ23x2=1-3x21-x23x4=1-3x21-x2
3x4=1-3x2+3x4-x24x2=1x2=14x=±12

Page No 4.92:

Question 4:

Prove that: cos-145+cos-11213=cos-13365

Answer:

L.H.S=cos-145+cos-11213=cos-145×1213-1-4521-12132          cos-1x+cos-1y=cos-1xy-1-x21-y2=cos-14865-35×513=cos-148-1565=cos-13365=R.H.S



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