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Page No 3.10:

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 3.10:

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 3.10:

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 3.10:

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 3.10:

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 3.115:

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 3.115:

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 3.115:

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 3.115:

Question 4:

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

(ii) sintan-11-x22x+cos-11-x21+x2=1
(iii) sin-12x1-x2=2cos1x,12x1
 

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
(iii)
To prove:
sin-12x1-x2=2cos-1x, 12x1
Let us consider cos-1x=θ
x=cosθ
Taking R.H.S.
sin-12cosθ1-cos2θsin-12cosθ sinθsin-1sin2θ   2sinθ cosθ=sin2θ2θ2cos-1x=RHS θ=cos-1x
Hence, proved.

Page No 3.115:

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 3.115:

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 3.116:

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 3.116:

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 3.116:

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 3.116:

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 3.116:

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 3.116:

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 3.116:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.117:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.118:

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 3.119:

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 3.119:

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 3.119:

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 3.119:

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 3.119:

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 3.119:

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 3.119:

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 3.119:

Question 36:

If α2 sin-1 x+cos-1 xβ, then(a) α=-π2, β=π2            (b) α=0, β=π            (c) α=-π2, β=3π2                (d) α=0, β=2π  

Answer:


2sin-1x+cos-1x

=sin-1x+sin-1x+cos-1x

=sin-1x+π2                              sin-1x+cos-1x=π2            

Now,

-π2sin-1xπ2

-π2+π2sin-1x+π2π2+π2

0sin-1x+π2π

02sin-1x+cos-1xπ

Comparing with α2sin-1x+cos-1xβ, we get

α=0,β=π

Hence, the correct answer is option (b).


 

Page No 3.119:

Question 37:

The value of sin (2sin-1(.6)) is
(a) 0.48            (b) 0.96             (c) 1.2            (d) sin 1.2

Answer:


sin2sin-10.6

=sinsin-12×0.6×1-0.62                2sin-1x=sin-12x1-x2

=sinsin-12×0.6×1-0.36

=sinsin-12×0.6×0.64

=sinsin-12×0.6×0.8

=sinsin-10.96

=0.96                              sinsin-1x=x,x-1,1

Hence, the correct answer is option (b).

Page No 3.119:

Question 38:

The value of cot (sin−1x) is
(a) 1+x2x             (b) x1+x2            (c) 1x                (d) 1-x2x

Answer:


We know

sin-1x=cot-11-x2xcotsin-1x=cotcot-11-x2xcotsin-1x=1-x2x                    cotcot-1x=x,xR

Thus, the value of cot(sin−1x) is 1-x2x.

Hence, the correct answer is option (d).

Page No 3.119:

Question 39:

If tan−1x10 for some x ∊ R, then the value of cot1 x is
(a) π5           (b) 2π5            (c) 3π5            (d) 4π5

Answer:


Disclaimer: The solution has been provided for the following question.

If tan−1π10 for some x ∊ R, then the value of cot1 x is
(a) π5           (b) 2π5            (c) 3π5            (d) 4π5

Solution:

We know

tan-1x+cot-1x=π2, xRπ10+cot-1x=π2                         tan-1x=π10cot-1x=π2-π10cot-1x=4π10=2π5

Hence, the correct answer is option (b).

Page No 3.119:

Question 40:

One branch of cos-1 other than the principal value branch corresponds to 
(a) π2,3π2            (b) π, 2π-3π2              (c) 0, π               (d) 2π, 3π

Answer:


The domain of the function fx=cos-1x is -1,1. The range of cos-1x in one of the intervals ...,-π,0,0,π,π,2π,2π,3π,... is one-one and onto with the range -1,1.

Thus, one branch of cos−1x other than the principal value branch corresponds to 2π,3π.

Hence, the correct answer is option (d).



Page No 3.120:

Question 41:

The principal value branch of sec-1 is
(a) -π2,π2-0              (b) 0, π-π2                (c) 0,π                (d) -π2,π2

Answer:


The principal value branch of sec−1x is 0, π-π2.

Hence, the correct answer is option (b).

Page No 3.120:

Question 42:

Which of the following corresponds to the principal value branch of tan-1?
​(a) -π2,π2              (b) -π2,π2             (c) -π2,π2-0                  (d) (0, π)

Answer:


The principal value branch of tan−1x is -π2,π2.

Hence, the correct answer is option (a).
 

Page No 3.120:

Question 43:

Which of the following is the principal value branch of cosec-1 ?
​(a) -π2,π2                (b) 0, π-π2                (c) -π2,π2                 (d) -π2,π2-0

Answer:


The principal value branch of cosec−1x is -π2,π2-0.

Hence, the correct answer is option (d).

Page No 3.120:

Question 44:

The value of the expression tan 12cos-125 is 
​(a) 2+5                (b) 5-2                (c) 5+22               (d) 5+2

Answer:


Let cos-125=θ. Then,
cosθ=25

Now,

tan12cos-125=tanθ2=1-cosθ1+cosθ
=1-251+25=5-25+2=5-225+25-2
=5-225-4=5-2

Thus, the value of the given expression is 5-2.

Hence, the correct answer is option (b).

Page No 3.120:

Question 45:

If 3 tan-1x + cot-1x = π, then x equals
​(a) 0            (b) 1           (c) -1           (d) 12

Answer:


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

Thus, the value of x is 1.

Hence, the correct answer is option (b).

Page No 3.120:

Question 46:

The value of sin-1 cos33π5 is 
​(a) 3π5             (b) -7π5               (c) π10                 (d) -π10

Answer:


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

Thus, the value of sin-1cos33π5 is -π10.

Hence, the correct answer is option (d).

Page No 3.120:

Question 47:

tan –1 3 + tan –1 λ= tan–1 3+λ1-3λ is valid for what value of λ
(a) λ=-13,13
(b) λ>13
(c) λ<13
(d) all real values of λ

Answer:

Given: tan –1 3 + tan –1 λ= tan–1 3+λ1-3λ
We know,
tan-1x+tan-1y=tan-1x+y1-xy if xy<13λ<1λ<13
Hence, the correct answer is option C.

Page No 3.120:

Question 48:

The value of tan–1tan7π6 is
(a) π6           (b) π2          (c)  π3        (d) 7π6

Answer:

We know,
tan-1tanx=x; if x t -π2,π2tan-1tan7π6=tan-1tanπ+π6=tan-1tanπ6   tanπ+θ=tanθ=π6   -π2<π6<π2 
Thus, the value of tan-1tan7π6 is π6
Hence, the correct answer is option A.
 

Page No 3.120:

Question 1:

The value of sec2(tan-12) + cosec2 (cot-1 3) is ____________________.

Answer:


We know

tan-1x=sec-11+x2 and cot-1x=cosec-11+x2

So,

tan-12=sec-11+22=sec-15

cot-13=cosec-11+32=cosec-110

sec2tan-12+cosec2cot-13

=sec2sec-15+cosec2cosec-110

=secsec-152+coseccosec-1102

=52+102

=15

Thus, the value of sec2tan-12+cosec2cot-13 is 15.

The value of sec2(tan−12) + cosec2(cot−1 3) is ____15____.

Page No 3.120:

Question 2:

If sin-1x - cos-1x = π6, then x = _________________________.

Answer:


Given: sin-1x-cos-1x=π6        .....(1)

We know

sin-1x+cos-1x=π2                   .....(2)

Adding (1) and (2), we get

2sin-1x=π6+π22sin-1x=4π6sin-1x=π3
x=sinπ3=32

If sin−1x − cos−1x = π6, then x =      32     .

Page No 3.120:

Question 3:

The range of sin-1x + cos-1x + tan-1x is _______________________.

Answer:


Domain of the given function = -1,1R = -1,1

Now,

For -1x1,

sin-1x+cos-1x=π2 and -π4tan-1xπ4

π2-π4sin-1x+cos-1x+tan-1xπ2+π4

π4sin-1x+cos-1x+tan-1x3π4

Thus, the range of the given function is π4,3π4.

The range of sin−1x + cos−1x + tan−1x is      π4,3π4    .

Page No 3.120:

Question 4:

If sin-1xπ5 for some x ∊ (-1, 1), then the value of cos-1 x is ____________________.

Answer:


Given: sin-1x=π5, x-1,1

We know

sin-1x+cos-1x=π2π5+cos-1x=π2cos-1x=π2-π5=3π10

Thus, the value of cos−1x is 3π10.

If sin−1xπ5 for some x ∈ (−1, 1), then the value of cos−1 x is      3π10     .



Page No 3.121:

Question 5:

If x < 0, then tan-1x + tan-11x is equal to ____________________.

Answer:


We know

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

tan-1x+tan-11x=tan-1x+cot-1x-π              x<0=π2-π                                 tan-1x+cot-1x=π2, xR=-π2

If x < 0, then tan−1x + tan−11x is equal to      -π2     .

Page No 3.121:

Question 6:

The value of tan-12 + tan-13 is ___________________.

Answer:


We know
tan-1x+tan-1y=π+tan-1x+y1-xy, if xy>1
tan-12+tan-13=π+tan-12+31-2 × 3=π+tan-1-1
=π-π4=3π4

The value of tan−12 + tan−13 is     3π4    .

Page No 3.121:

Question 7:

If tan-1 -13 + cot-1 x = x2, then x = __________________.

Answer:


Disclaimer: The solution is provided for the following question.

If tan−1 -13 + cot1 x = π2, then x = __________________.

Solution:

We know

tan-1x+cot-1x=π2, for all x ∈ R

tan-1-13+cot-1-13=π2       .....(1)

It is given that,

tan-1-13+cot-1x=π2                        .....(2)

From (1) and (2), we get

x=-13

If tan−1 -13 + cot1 x = π2, then x =      -13     .

Page No 3.121:

Question 8:

If tan-1 x-tan-1 yπ4, then x - y - xy = ____________________.

Answer:


tan-1x-tan-1y=π4tan-1x-y1+xy=π4x-y1+xy=tanπ4
x-y1+xy=1x-y=1+xyx-y-xy=1

If tan−1  tan1 y =  π4, then x y xy = __1__.

Page No 3.121:

Question 9:

The value of cot (tan-1x + cot-1x) for all x ∊ R, is ____________________

Answer:


We know

tan-1x+cot-1x=π2, for all x ∈ R

cottan-1x+cot-1x=cotπ2

cottan-1x+cot-1x=0

The value of cot(tan−1x + cot−1x) for all x ∈ R, is __0__.

Page No 3.121:

Question 10:

If cos-1x + cos-1 y = π3, then sin-1 x + sin-1 y =____________________.

Answer:


We know

sin-1x+cos-1x=π2, for all x ∈ R                  .....(1)

Also, sin-1y+cos-1y=π2, for all y ∈ R        .....(2)

Adding (1) and (2), we get

sin-1x+cos-1x+sin-1y+cos-1y=π2+π2

sin-1x+sin-1y+cos-1x+cos-1y=π

sin-1x+sin-1y+π3=π                 (Given)

sin-1x+sin-1y=π-π3=2π3

If cos−1 x + cosy = π3, then sinx + siny =      2π3     .

Page No 3.121:

Question 11:

If x > 0, y > 0, xy > 1, then tan-1x + tan-1y = _____________________.

Answer:


We know

tan-1x+tan-1y=π+tan-1x+y1-xy, if x > 0, y > 0 and xy > 1

If x > 0, y > 0, xy > 1, then tan−1x + tan−1y =      π+tan-1x+y1-xy     .
 

Page No 3.121:

Question 12:

If 3 sin-1x = π-cos-1x, then x = __________________.

Answer:


3sin-1x=π-cos-1x3π2-cos-1x=π-cos-1x               sin-1x+cos-1x=π23π2-3cos-1x=π-cos-1x
2cos-1x=3π2-π=π2cos-1x=π4x=cosπ4=12

If 3sin−1x = π cos1x, then x =      12     .
 

Page No 3.121:

Question 13:

If tan-1x + tan-1 y = 5π6, then cot-1x + cot-1y = _________________.

Answer:


We know

tan-1a+cot-1a=π2, for all a ∈ R        .....(1)

Now,

tan-1x+tan-1y=5π6           (Given)

π2-cot-1x+π2-cot-1y=5π6             [Using (1)]

cot-1x+cot-1y=π-5π6

cot-1x+cot-1y=π6

If tan−1x + tan−1 y = 5π6, then cot−1+ cot−1y =      π6     .

Page No 3.121:

Question 14:

If tan-1x - cot-1x = tan-13, then x = _______________________.

Answer:


tan-1x-cot-1x=tan-13tan-1x-π2-tan-1x=π3                  tan-1x+cot-1x=π22tan-1x=π2+π3tan-1x=π4+π6
x=tanπ4+π6x=tanπ4+tanπ61-tanπ4×tanπ6x=1+131-13=3+13-1
x=3+123-13+1x=4+232x=2+3

If tan−1x  cot1x = tan13, then x =      2+3     .

 

Page No 3.121:

Question 15:

If sin-1x + sin-1y + sin-1z = -3π2, then xyz = __________________.

Answer:


We know

-π2sin-1aπ2, for all a ∈ [−1, 1]

So, the minimum value of sin−1a is -π2.

Now,

sin-1x+sin-1y+sin-1z=-3π2     (Given)

This is possible if

sin-1x=-π2, sin-1y=-π2 and sin-1z=-π2

x = −1, y = −1 and z = −1

xyz = (−1) × (−1) × (−1) =  −1

If sin−1x + sin−1y + sin−1z = -3π2, then xyz = ___−1___.

Page No 3.121:

Question 16:

The value of cos-1 sincos-112is ________________________.

Answer:


cos-1sincos-112=cos-1sinπ3                 cosπ3=12cos-112=π3=cos-132=π6                                        cosπ6=32cos-132=π6

Thus, the value of cos-1sincos-112 is π6.

The value of cos-1sincos-112 is     π6    .

Page No 3.121:

Question 17:

The value of tan cos-1sincot-11 is ___________________.

Answer:


tancos-1sincot-11=tancos-1sinπ4                cotπ4=1cot-11=π4=tancos-112
=tanπ4                                            cosπ4=12cos-112=π4=1

Thus, the value of tancos-1sincot-11 is 1.

The value of tancos-1sincot-11 is __1__.

Page No 3.121:

Question 18:

The value of tan​(sec-13) + cot2 (cosec-14) is _________________.

Answer:


tan2sec-13+cot2cosec-14=sec2sec-13-1+cosec2cosec-14-1                       1+tan2θ=sec2θ and1+cot2θ=cosec2θ=secsec-132+coseccosec-142-2
=32+42-2=9+16-2=23

Thus, the value of tan2sec-13+cot2cosec-14 is 23.

The value of tan​(sec−13) + cot2 (cosec−14) is ____23____.

Page No 3.121:

Question 19:

If tan−1(cotθ) = 2θ, then θ = __________________.

Answer:


tan-1cotθ=2θtan-1tanπ2-θ=2θπ2-θ=2θ3θ=π2θ=π6

Thus, the value of θ is π6.

If tan−1(cotθ) = 2θ, then θ =      π6     .

Page No 3.121:

Question 20:

The value of sin-1 cos33π5 is _________________.

Answer:


sin-1cos33π5=sin-1cos6π+3π5=sin-1cos3π5
=sinsinπ2-3π5=sinsin-π10=-π10

Thus, the value of sin-1cos33π5 is -π10.

The value of sin−1cos33π5 is      -π10     .

Page No 3.121:

Question 21:

If tan-1x + tan-1y = 4π5, then cot-1x + cot-1y = _________________.

Answer:


We know

tan-1a+cot-1a=π2, for all a ∈ R          .....(1)

Now,

tan-1x+tan-1y=4π5              (Given)

π2-cot-1x+π2-cot-1y=4π5            [Using (1)]

cot-1x+cot-1y=π-4π5

cot-1x+cot-1y=π5

If tan−1x + tan−1y = 4π5, then cot−1x + cot−1y =      π5     .

Page No 3.121:

Question 22:

If 3 tan-1x + cot-1x = π, then x = ____________________.

Answer:


3tan-1x+cot-1x=π              (Given)

2tan-1x+π2=π                               tan-1x+cot-1x=π2

2tan-1x=π-π2=π2

tan-1x=π4

x=tanπ4=1

Thus, the value of x is 1.

If 3tan−1x + cot−1x = π, then x = ___1___.

 

Page No 3.121:

Question 23:

If tan-12, tan-13 are measures of two angles of triangle, then the measure of its third angle is _________________.

Answer:


Let the measure of third angle of the triangle be x.

Now,

tan-12+tan-13+x=π                           (Angle sum property of triangle)

π+tan-12+31-2×3+x=π              tan-1x+tan-1y=π+tanx+y1-xy, if xy>1

tan-1-1+x=0

-π4+x=0

x=π4

Thus, the measure of third angle of the triangle is π4.

If tan−12, tan−13 are measures of two angles of triangle, then the measure of its third angle is      π4     .

Page No 3.121:

Question 24:

If tan-1ax+ tan-1bx=π2, then x = _________________.

Answer:


We know

tan-1x=cot-11x

tan-1bx=cot-11bx=cot-1xb              .....(1)
So,

tan-1ax+tan-1bx=π2       (Given)

tan-1ax+cot-1xb=π2                      [Using (1)]

ax=xb                        tan-1y+cot-1y=π2

x2=ab

x=ab

If tan−1ax+ tan−1bx=π2, then x =      ab     .

 

Page No 3.121:

Question 25:

If cos(2sin-1x) = 19, then the value of x is ______________.

Answer:


Let sin-1x=θsinθ=x.

cos2sin-1x=19cos2θ=191-2sin2θ=19
2sin2θ=1-19=89sin2θ=49sinθ=±23x=±23                        sinθ=x

Thus, the value of x is ±23.

If cos(2sin−1x) = 19, then the value of x is      ±23     .

Page No 3.121:

Question 26:

If 0 < < π2, then sin-1 (cos x) + cos-1 (sin x) = ___________________.

Answer:


sin-1cosx+cos-1sinx

=sin-1sinπ2-x+cos-1cosπ2-x

=π2-x+π2-x

=π-2x

If 0 < π2, then sin−1(cos x) + cos−1(sin x) =      π-2x     .

Page No 3.121:

Question 27:

If tan-1x=π4-tan-113, then x = ______________________.

Answer:


tan-1x=π4-tan-113tan-1x=tan-11-tan-113tan-1x=tan-11-131+1×13                     tan-1x-tan-1y=tan-1x-y1+xy, xy>-1
tan-1x=tan-12343tan-1x=tan-112x=12

If tan-1x=π4-tan-113, then x =      12     .




 

Page No 3.121:

Question 28:

If tan-1x + tan-1 12=π4, then x = _________________.

Answer:


tan-1x+tan-112=π4tan-1x + 121-x × 12=π4                                tan-1a+tan-1b=tan-1a+b1-ab2x+12-x=tanπ4=1
2x+1=2-x3x=1x=13

Thus, the value of x is 13.

If tan−1x + tan−112=π4, then x =      13     .

Page No 3.121:

Question 29:

cot π4-2cot-13 is equal to ______________________.

Answer:


Let cot-13=θcotθ=3.

cotπ4-2cot-13=cotπ4-2θ=cotπ4cot2θ+1cot2θ-cotπ4              cotA-B=cotAcotB+1cotB-cotA
=cot2θ+1cot2θ-1=cot2θ-12cotθ+1cot2θ-12cotθ-1=cot2θ-1+2cotθcot2θ-1-2cotθ
=32-1+2×332-1-2×3                    cotθ=3=9-1+69-1-6=142=7

cot π4-2cot-13 is equal to ___7___.

Page No 3.121:

Question 30:

tan-1 tan2π3is equal to __________________.

Answer:


tan-1tan2π3=tan-1tanπ-π3=tan-1-tanπ3
=tan-1tan-π3                           tan-θ=-tanθ=-π3                                                tan-1tanx=x, if x-π2,π2

tan−1tan2π3 is equal to      -π3     .

Page No 3.121:

Question 31:

If y = 2tan-1 x+sin-12x1+x2for all x, then y lies in the interval_________________.

Answer:


We know

2tan-1x=sin-12x1+x2,-1x1π-sin-12x1+x2,x>1-π-sin-12x1+x2,x<-1

y=2tan-1x+sin-12x1+x2=4tan-1x,-1x1π,x>1-π,x<-1

For -1x1,

-π4tan-1xπ4-π4tan-1xπ-πyπ                .....1

For x > 1, y = π             .....(2)

For x < −1, y = -π        .....(3)

From (1), (2) and (3), we get

y-π,π, for all x ∈ R

Thus, the range of y is -π,π.

If y = 2tan−1x + sin−12x1+x2 for all x, then y lies in the interval      -π,π     .



Page No 3.122:

Question 32:

The result tan-1 x-tan-1 y = tan-1x-y1+xy is true when value of xy is __________________.

Answer:


We know

tan-1x-tan-1y=tan-1x-y1+xy,if xy>-1π+tan-1x-y1+xy,if x>0,y<0,xy<-1-π+tan-1x-y1+xy,if x<0,y>0,xy<-1

Thus, tan-1x-tan-1y=tan-1x-y1+xy when the value of xy > −1.

The result tan−1x − tan−1y = tan−1x-y1+xy is true when value of xy is __greater than − 1___.

Page No 3.122:

Question 33:

The value of cot-1(-x) for all x ∊ R in terms of cot-1 x is _________________.

Answer:


We know

cot-1-x=π-cot-1x, for all x ∈ R

The value of cot−1(−x) for all x ∈ R in terms of cot−1 x is      π-cot-1x     .

Page No 3.122:

Question 34:

The principal values of cos-1-12 is ..................

Answer:

We know, cos-1-θ=π-cos-1θ
cos-1-12=π-cos-112=π-π3 cos-112=π3=2π3
Thus, the principal value of cos-1-12 is 2π3

Page No 3.122:

Question 35:

The range of the principal value branch of y = sec–1 x is ........................

Answer: