Rd Sharma Xi (2018) Solutions for Class 12 Science Math Chapter 5 Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Trigonometric Functions are extremely popular among class 12 Science students for Math Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi (2018) Book of class 12 Science Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi (2018) Solutions. All Rd Sharma Xi (2018) Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 5.18:

Question 1:

stion Prove the following identities (1-16)
sec4 x sec2 x = tan4 x + tan2 x

Answer:

LHS = sec4 x-sec2 x         =sec2 xsec2 x-1         =tan2 x+1tan2 x             sec2 x-tan2 x = 1         =tan4 x+tan2 x         =RHSHence proved.

Page No 5.18:

Question 2:

Prove the following identities (1-16)
sin6 x +cos6 x=1-3 sin2 x cos2 x

Answer:

LHS = sin6x+cos6x         =sin2x3+cos2x3         =sin2x+cos2xsin2x2+cos2x2-sin2x cos2x        a3 + b3 = a+ba2+b2-ab         =1×sin2x+cos2x2-2sin2x cos2x-sin2x cos2x        sin2x+cos2x = 1 and a2 + b2 = a+b2-2ab         =12 - 3 sin2x cos2x         =1-3 sin2x cos2x         =RHSHence proved.

Page No 5.18:

Question 3:

Prove the following identities (1-16)

cosec x-sin x sec x-cos x tan x+cot x=1

Answer:

LHS = cosec x-sin x sec x-cos x tan x+cot x         = 1sin x-sin x 1cos x-cos xsin xcos x+cos xsin x          =1-sin2xsin x1-cos2xcos xsin2 x+cos2xcos x sin x         =cos2xsin xsin2 xcos x1cos x sin x         =1         =RHSHence proved.

Page No 5.18:

Question 4:

Prove the following identities (1-16)
cosec x sec x-1-cot x 1-cos x=tan x-sin x

Answer:

LHS = cosec x sec x-1-cot x 1-cos x         =1sin x 1cos x-1-cos xsin x 1-cos x         =1sin x 1-cos xcos x-cos xsin x 1-cos x         = 1-cos xsin x1cos x-cos x         = 1-cos xsin x1-cos2 xcos x         =1-cos xsin xsin2 xcos x         =1-cos xsin xcos x        =sin xcos x-sin x        =tan x-sin x        = RHSHence proved.

Page No 5.18:

Question 5:

Prove the following identities (1-16)
1-sin x cos xcos x sec x-cosec x·sin2 x-cos2 xsin3 x+cos3 x=sin x

Answer:

LHS=1-sin x cos xcos x sec x-cosec x×sin2 x-cos2 xsin3 x+cos3 x       =1-sin x cos xcos x 1cos x-1sin x×sin x2-cos x2sin x3+cos x3       =1-sin x cos xcos x sin x-cos xcos x sin x×sin x+cos xsin x-cos xsin x+cos xsin x2+cos x2-sin x cos x       = sin x1-sin x cos x sin x-cosx×sin x+cos xsin x-cos xsin x+cos xsin x2+cos x2-sin x cos x       = sin x1-sin x cos x 1×1×1sin2 x+cos2 x-sin x cos x       =sin x1-sin x cos x×11-sinx cos x       =sin x       =RHSHence proved.

Page No 5.18:

Question 6:

Prove the following identities (1-16)

tan x1-cot x+cot x1-tan x=sec x cossec x+1

Answer:

LHS = tan x1-cot x+cot x1-tan x         =sin xcos x1-cos xsin x+cos xsin x1-sin xcos x         =sin xcos xsin x-cos xsin x+cos xsin xcos x-sin xcos x         =sin xcos x×sin xsin x-cos x+cos xsin x×cos xcos x-sin x         =sin xcos x×sin xsin x-cos x+cos xsin x×cos x-sin x-cos x         =sin2 xcos xsin x-cos x-cos2xsin xsin x-cos x         =sin3 x-cos3xsin x cos xsin x-cos x         =sin x-cos xsin2 x+cos2 x+sin xcos xsin x cos xsin x-cos x         =1×1+sin x cos xsin x cos x         =1+sin x cos xsin x cos x         =1sin x cos x+sin x cos xsin x cos x         =1sin x ×1cos x+1         =cosec x ×sec x+1         =sec x cosec x+1         =RHSHence proved.

Page No 5.18:

Question 7:

Prove the following identities (1-16)
sin3 x+cos3 xsin x+cos x+sin3 x-cos3 xsin x-cos x=2

Answer:

LHS=sin3 x+cos3 xsin x+cos x+sin3 x-cos3 xsin x-cos x        =sin x+cos xsin2 x+cos2 x - sin x cos xsin x+cos x+sin x-cos xsin2 x+cos2 x + sin x cos xsin x-cos x        =sin2 x+cos2 x - sin x cos x + sin2 x+cos2 x + sin x cos x        =1 - sin x cos x +1 + sin x cos x        =2       = RHSHence proved.

Page No 5.18:

Question 8:

Prove the following identities (1-16)
sec x sec y+tan x tan y2-sec x tan y+tan x sec y2=1

Answer:

LHS = sec x sec y+tan x tan y2-sec x tan y+tan x sec y2         =sec x sec y2+tan x tan y2-2sec x sec ytan x tan y               -  sec x tan y2+tan x sec y2-2sec x tan ytan x sec y          =sec2 x sec2 y+tan2 x tan2 y-2sec x sec y tan x tan y               -  sec2 x tan2 y+tan2 x sec2 y-2 sec x sec y tan x tan y         =sec2 x sec2 y+tan2 x tan2 y-2 sec x sec y tan x tan y               -  sec2 x tan2 y-tan2 x sec2 y+2 sec x sec y tan x tan y         =sec2 x sec2 y-  sec2 x tan2 y+tan2 x tan2 y-tan2 x sec2 y         =sec2 x sec2 y- tan2 y+tan2 x tan2 y- sec2 y         =sec2 x sec2 y- tan2 y-tan2 x sec2 y- tan2 y         =sec2 x×1-tan2 x×1         =sec2 x-tan2 x         =1        =RHSHence proved.

Page No 5.18:

Question 9:

Prove the following identities (1-16)
cos x1-sin x=1+cos x+sin x1+cos x-sin x

Answer:

RHS=1+cos x+sin x1+cos x-sin x        =1+cos x+sin x1+cosx-sin x        =1+cos x+sin x1+cos x+sin x1+cos x-sin x1+cos x+sin x        =1+cos x+sin x21+cos x2-sin x2        =1+cos x2+sin x2+21+cos xsin x12+cos2 x+2×1×cos x-sin2 x        =1+cos2 x+2cos x+sin2 x+2sin x cos x+2sin x1+cos2 x+2cos x-sin2 x        =1+sin2 x+cos2 x+2cos x+2sin x cos x+2sin x1-sin2 x+cos2 x+2cos x        =1+1+2cos x+2sin x cos x+2sin xcos2 x+cos2 x+2cos x        =2+2cos x+2sin x cos x+2sin x2cos2 x+2cos x        =1+cos x+sin x cos x+sin xcos2 x+cos x        =11+cos x+sin x cos x+1cos xcos x+1        =cos x+11+sin xcos xcosx+1        =1+sin xcos x        =1+sin x×cos xcos x×cos x        =1+sin x×cos xcos2 x        =1+sin x×cos x1-sin2 x        =1+sin x×cos x1+sin x1-sin x        =cos x1-sin x        =LHSHence proved.

Page No 5.18:

Question 10:

Prove the following identities (1-16)

tan3 x1+tan2 x+cot3 x1+cot2 x=1-2 sin2 x cos2 xsin x cos x

Answer:

LHS = tan3 x1+tan2 x+cot3 x1+cot2 x        =tan3 xsec2 x+cot3 xcosec2 x        =sin3 xcos3 x1cos2 x+cos3 xsin3 x1sin2 x        =sin3 xcos3 x×cos2 x1+cos3 xsin3 x×sin2 x1        =sin3 xcos x+cos3 xsin x        =sin4 x+cos4 xsin x cos x        =sin2 x2+cos2 x2sin x cos x        =sin2 x+cos2 x2-2sin2 x cos2 xsin x cos x        =12-2sin2 x cos2 xsin x cos x        =1-2sin2 x cos2 xsin x cos x        =RHSHence proved.

Page No 5.18:

Question 11:

Prove the following identities (1-16)

1-sin2 x1+cot x-cos2 x1+tan x=sin x cos x

Answer:

1-sin2 x1+cot x-cos2 x1+tan x=sin x cos x


LHS= 1-sin3xsinx+cosx-cos3xsinx+cosx=sinx+cosx-sin3x+cos3xsinx+cosx=sinx+cosx1-sin2x-cos2x+sinx cosxsinx+cosx=1-sin2x-cos2x+sinx cosx=1-1+sinx cosx=sinx cosx

= RHS
Hence proved.

Page No 5.18:

Question 12:

Prove the following identities (1-16)
1sec2 x-cos2 x+1cosec2 x-sin2 x sin2 x cos2 x=1-sin2 x cos2 x2+sin2 x cos2 x

Answer:

1sec2 x-cos2 x+1cosec2 x-sin2 x sin2 x cos2 x=1-sin2 x cos2 x2+sin2 x cos2 x

LHS =1sec2x-cos2x+1cosec2x-sin2xsin2x cos2x=11cos2x-cos2x+11sin2x-sin2xsin2x cos2x=cos2x1-cos4x+sin2x1-sin4xsin2x cos2x=cos2x1-sin4x+sin2x1-cos4x1-cos4x1-sin4xsin2x cos2x=1-cos2x sin4x-cos4x sin2x1+sin2x1-sin2x1+cos2x1-cos2xsin2x cos2x=1-cos2x sin4x-cos4x sin2x1+sin2x.cos2x.1+cos2x.sin2xsin2xcos2x=1-cos2xsin4x-cos4xsin2x1+sin2x1+cos2x =1-cos2xsin2xsin2x+cos2x2+sin2x.cos2x=1-cos2x sin2x2+sin2x.cos2x

= RHS

Hence proved.

Page No 5.18:

Question 13:

Prove the following identities (1-17)

1+tan α tan β2+tan α-tan β2=sec2 α sec2 β

Answer:

1+tan α tan β2+tan α-tan β2=sec2 α sec2 β

LHS = 1+tanα tanβ2+tanα-tanβ2        =1+tan2α tan2β+2tanα tanβ+tan2α+tan2β-2tanα tanβ        =1+tan2α tan2β+tan2α+tan2β        =tan2αtan2β+1+11+tan2β        =1+tan2β1+tan2α        =sec2α.sec2β         = RHS

Hence proved.

Page No 5.18:

Question 14:

Prove the following identities (1-16)

1+cot x+tan x  sin x-cos xsec3 x-cosec3 x=sin2 x cos2 x

Answer:

1+cot x+tan x  sin x-cos xsec3 x-cosec3 x=sin2 x cos2 x

LHS=1+cotx+tanxsinx-cosxsec3x-cosec3x        =1+cosxsinx+sinxcosxsinx-cosx1cos3x-1sin3x        =sinx cosx+cos2x+sin2xsinx-cosxsin2xcos2xsin3x-cos3x        =1+sinx cosxsinx-cosxsin2x cos2xsinx-cosxsin2x+cos2x+sinx cosx        =sin2x cos2x         =RHSHence proved.

Page No 5.18:

Question 15:

Prove the following identities (1-16)

2 sin x cos x-cos x1-sin x+sin2 x-cos2 x=cot x

Answer:

LHS = 2sinx cosx-cosx1-sinx+sin2x-cos2x        =cosx(2sinx-1)2sin2x-sinx           1-cos2x=sin2x        =cosx(2sinx-1)sinx2sinx-1        =cotx          =RHSHence proved.

Page No 5.18:

Question 16:

Prove the following identities (1-16)

cos x tan x+2 2 tan x+1=2 sec x+5 sin x

Answer:

LHS = cosxtanx+22tanx+1        =cosx2tan2x+5tanx+2        =cosx2sin2xcos2x+5sinxcosx+2        =2sin2x+5sinx cosx+2cos2xcosx        =2+5sinx cosxcosx        =2secx+5sinx         = RHSHence proved.

Page No 5.18:

Question 17:

If x=2 sin x1+cos x+sin x, then prove that 1-cos x+sin x1+sin x is also equal to a.

Answer:

Disclaimer: There is some error in the given question.
The question should have been
Question: If a=2 sin x1+cos x+sin x, then prove that 1-cos x+sin x1+sin x is also equal to a.
So, the solution is done accordingly.

Solution:
a = 2sinx1+sinx+cosxRationalising the denominator:2sinx1+sinx+cosx×1+sinx-cosx1+sinx-cosx=2sinx1+sinx-cosx1+sinx2-cos2x=2sinx1+sinx-cosx1+sin2x+2sinx-cos2x=2sinx1+sinx-cosx2sin2x+2sinx=2sinx1+sinx-cosx2sinx1+sinx=1+sinx-cosx1+sinx a=1+sinx-cosx1+sinx

Hence proved.

Page No 5.18:

Question 18:

If sin x=a2-b2a2+b2, then the values of tan x, sec x and cosec x

Answer:

 sinx=a2-b2a2+b2We know:sin2x+cos2x=1cos2x=1-sin2x          =1-a2-b2a2+b22          =a4+b4+2a2b2-a4+b4-2a2b2a2+b22          =4a2b2a2+b22cosx = 2aba2+b2

tanx=sinxcosx=a2-b2a2+b22aba2+b2=a2-b22absecx=1cosx=a2+b22abcosecx=1sinx=a2+b2a2-b2

Page No 5.18:

Question 19:

If tan x=ba, then find the values of a+ba-b+a-ba+b.

Answer:

tanx=baNow, a+ba-b+a-ba+b=1+ba1-ba+1-ba1+ba=1+tanx1-tanx+1-tanx1+tanx=tanx+1+1-tanx1-tan2x=21-tan2x=2cosxcos2x-sin2x

Page No 5.18:

Question 20:

If tan x=ab, show that a sin x-b cos xa sin x+b cos x=a2-b2a2+b2.

Answer:

LHS:asinx-bcosxasinx+bcosxDividing by bcosx:= atanxb-1atanxb+1Substituting the value of tanx= a2-b2a2+b2= RHS
Hence proved.



Page No 5.19:

Question 21:

If cosec x-sin x=a3, sec x-cos x=b3, then prove that a2 b2 a2+b2=1.

Answer:

 cosec x-sin x=a31sin x-sin =a31-sin2xsin x=a3cos2xsin x=a3     a=cos2xsin x13     ....(i)Also, sec x-cos x=b31cos x-cos =b31-cos2xcos x=b3sin2xcos x=b3   b= sin2xcos x13         .....(ii)Now, LHS=a2b2a2+b2 =ab2a2+b2=cos2xsin x13sin2xcos x132 cos2xsin x132+sin2xcos x132=sin x cos x23 cos2x23sin x23+sin2x23cos x23=sin x cos x23cos3x23+sin3x23sin x23cos x23=sin x cos x23cos2x+sin2xsin x cos x23=1=RHS

Page No 5.19:

Question 22:

If cot x 1+sin x=4 m and cot x 1-sin x=4 n, prove that m2+n22=mn.

Answer:

Given:4m=cotx1+sinx and 4n=cotx1-sinxMultiplying both the equations:16mn=cot2x1-sin2x16mn=cot2x.cos2xmn=cos4x16 sin2x                  1Squaring the given equation:16m2=cot2x1+sinx2 and 16n2=cot2x1-sinx216m2-16n2=cot2x4sinxm2-n2=cot2x.sinx4Squaring both sides,m2-n22=cot4x.sin2x16m2-n22=cos4x16sin2x           (2)From (1) and (2):m2-n22= mnHence proved.

Page No 5.19:

Question 23:

If sin x+cos x=m, then prove that sin6 x+cos6 x=4-3 m2-124, where m22

Answer:

sin x+ cos x =m         (Given)To prove: sin6 x +cos6 x =4-3 (m2-1)24  , where m22Proof:LHS:  sin6 x +cos6 x =sin2 x3+cos2 x3=sin2 x+cos2 x 3-3sin2 x cos2 xsin2 x+cos2 x=1-3sin2 x cos2 xRHS: 4-3 (m2-1)24   =4-3sin x+ cos x 2-124=4-3sin2x+ cos2 x+2sinx cosx-124=4-3sin2x-1-cos2 x+2 sinx cosx24=4-3×4 sin2x cos2 x4=1-3 sin2 x cos2 xLHS=RHSHence proved

Page No 5.19:

Question 24:

If a=sec x-tan x and b=cosec x+cot x, then shown that ab+a-b+1=0.

Answer:

a=secx-tanx      And, b=cosecx+cotx   =1-sinxcosx          And, b=1+cosxsinxNow, we have:ab+a-b+11-sinxcosx1+cosxsinx+1-sinxcosx-1+cosxsinx+1=1-sinx+cosx-sinx cosx+sinx-sin2x-cosx-cos2x+sinx cosxsinx cosx=1-sin2x-cos2xsinx cosx=0

Hence proved.

Page No 5.19:

Question 25:

Prove the: 1-sin x1+sin x+1+sin x1-sin x=-2cos x, whereπ2<x<π

Answer:

LHS = 1-sin x1+sin x+1+sin x1-sin x         = 1-sin x1-sin x1+sin x1-sin x+1+sin x1+sin x1-sin x1+sin x         = 1-sin x1-sin x1+sin x1-sin x+1+sin x1+sin x1-sin x1+sin x         = 1-sin x21-sin2 x+1+sin x21-sin2 x         =1-sin x2cos2 x+1+sin x2cos2 x         =1-sin xcos x+1+sin xcos x         =1-sin x+1+sin xcos x         =2cos x         =-2cos x         π2<x<π and in the second quadrant, cosx is negative         =RHSHence proved.

Page No 5.19:

Question 26:

If Tn=sinn x+cosn x, prove that

(i) T3-T5T1=T5-T7T3

(ii) 2 T6-3 T4+1=0

(iii) 6T10-15 T8+10 T6-1=0

Answer:

(i) LHS:

T3-T5T1=sin3x+cos3x-sin5x+cos5xsinx+cosx= sin3x-sin5x+cos3x-cos5xsinx+cosx=sin3x1-sin2x+cos3x1-cos2xsinx+cosx=sin3x.cos2x+cos3x.sin2xsinx+cosx=sin2x.cos2xsinx+cosxsinx+cosx=sin2x.cos2x

RHS:
T5-T7T3=sin5x+cos5x-sin7x+cos7xsin3x+cos3x=sin5x-sin7x+cos5x-cos7xsin3x+cos3x=sin5x1-sin2x+cos5x1-cos2xsin3x+cos3x=sin5xcos2x+cos5xsin2xsin3x+cos3x=sin2x.cos2x

LHS = RHS

Hence proved.

(ii) LHS:

2T6-3T4+12sin6x+cos6x-3sin4x+cos4x+12sin2x+cos2xsin4x+cos4x-sin2x cos2x-3sin4x+cos4x+12.1.sin4x+cos4x-sin2x cos2x-3sin4x+cos4x+12sin4x+2cos4x-2sin2x cos2x-3sin4x-3cos4x+1-sin4x+cos4x-sin2x cos2x+1-(sin2x+cos2x)2+1-1+10

Hence proved.

(iii) LHS:

6T10-15T8+10T6-16sin10x+cos10x-15sin8x+cos8x+10sin6x+cos6x-1



Page No 5.25:

Question 1:

Find the values of the other five trigonometric functions in each of the following:

(i) cot x=125, x in quadrant III

(ii) cos x=-12, x in quadrant II

(iii) tan x=34, x in quadrant III

(iv) sin x=35, x in quadrant I

Answer:

i We have:cotx = 125 and x are in the third quadrant. In the third quadrant, tanx and cotx are positiveAnd, sinx, cosx , secx and cosecx are negative.
 tanx = 1cotx = 1125=512cosecx = -1 + cot2x = -1 +1252 = -135sinx = 1cosecx = 1- 135 = -513cosx = -1 - sin2x = -1 - -5132 = -1213And, secx = 1cosx = 1-1213 = -1312
ii We have:cosx =-12 and x are in the second quadrant.In the second quadrant, sinx, and cosecx are positive.And, tanx, cotx , cosx and secx are negative.

 sinx = 1 - cos2x = 1 - -122 = 32tanx = sinxcosx = 32-12=-3
cotx= 1tanx = 1-3=-13 secx = 1cosx = 1-12 = -2cosecx = 1sinx = 132 = 23
iii We have:tanx = 34 and x are in the third quadrant.In the third quadrant, tanx, and cotx are positive.And, sinx, cosx , secx and cosecx are negative.
 cotx = 1tanx = 134=43secx = -1 +tan2x = -1 +342 = - 54cosx = 1secx = 1- 54 = -45sinx = -1 - cos2x = -1 - -452 = -35cosecx = 1sinx = 1- 35 = -53
iv We have:sinx =35 and x are in the first quadrant.In the first quadrant,  all six T-ratios are positive.
 cosx = 1 - sin2x = 1 - 352 = 45tanx = sinxcosx = 3545=34cotx = 1tanx = 134=43secx = 1cosx = 145 = 54cosecx = 1sinx = 135 = 53

Page No 5.25:

Question 2:

If sin x=1213 and x lies in the second quadrant, find the value of sec x + tan x.

Answer:

We have:sinx = 1213 and x lie in the second quadrant.In the second quadrant, sinx and cosecx are positive and all the other four T-ratios are negative.
 cosx = -1-sin2x               =-1-12132                = -513tanx = sinxcosx          = 1213-513         =-125And, sec x = 1cos x                     =1-513                    = -135 secx + tanx = -135 + -125                             = -5

Page No 5.25:

Question 3:

If sin x=35, tan y=12 and π2<x<π<y<3π2, find the value of 8 tan x-5 sec y.

Answer:

We have:
sinx=35, tany=12 and π2<x<π<y<3π2,Thus, x is in the second quadrant and y is in the third quadrant.In the second quadrant, cos x and tan x are negative.In the third quadrant, secy is negative.
 cosx = -1-sin2 x = -1-352 = -45tanx = 35-45 = -34And, secy = -1+tan2y = -1+122 = -52 8 tanx - 5 secy = 8×-34 - 5×-52 =-6 + 52=-72

Page No 5.25:

Question 4:

If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.

Answer:

We have:sinx + cosx = 0 sinx = -cosx sinxcosx = -1 tanx = -1

Now, x is in the fourth quadrant.In the fourth quadrant, cosx and sec x are positive and all the other four T-ratios are negative.

 secx = 1+tan2x = 1+-12 = 2cosx = 1secx = 12And, sinx = -1-cos2x = -1-122 = -12  sinx =-12 and cosx  = 12

Page No 5.25:

Question 5:

If cosx=-35 and π<x<3π2 find the values of other five trigonometric functions and hence evaluate cosec x+cot xsec x-tan x.

Answer:

We have:cos x = -35 and π<x<3π2Thus, x is in the third quadrant.In the third quadrant, tanx and cotx are positiveAnd, all the other four T-ratios are negative. sinx = -1 - cos2x = -1 - -352 = -45tanx = sin xcos x =  -45-35=43cot x = 1tan x = 143= 34secx = 1cos x=1-35 = -53cosec x = 1sin x = 1-45 = -54

Now, cosecθ + cotθsecθ-tanθ = -54+34-53-43                              =-24-93                              =-12-3 = 16  



Page No 5.39:

Question 1:

Find the values of the following trigonometric ratios:
(i) sin5π3

(ii) sin 17π

(iii) tan11π6

(iv) cos-25π4

(v) tan 7π4

(vi) sin17π6

(vii) cos19π6

(viii) sin-11π6

(ix) cosec-20π3

(x) tan-13π4

(xi) cos19π4

(xii) sin41π4

(xiii) cos39π4

(xiv) sin151π6

Answer:

i We have:5π3 = 53×180° = 300° = 90°×3 + 30°300° lies in the fourth quadrant in which the sine function is negative.Also, 3 is an odd integer. sin5π3 = sin 300° =sin 90°×3 + 30° = -cos30° = -32

ii We have:sin 17π=sin 3060°3060° = 90°×34 + 0°Clearly 3060° is in the negative direction of the x-axis, i.e. on the boundary line of the II and III quadrants.Also, 34 is an even integer. sin3060° =sin 90°×34 + 0° =-sin 0° =0 

iii We have:11π6 = 116×180° = 330° = 90°×3 + 60°330° lies in the fourth quadrant in which the tangent function is negative.Also, 3 is an odd integer. tan11π6 = tan 330° =tan90°×3 + 60° = -cot 60° = -13

iv We have:cos-25π4=cos1125°cos -1125° =cos 1125°=  cos 90°×12 + 45°1125° lies in the first quadrant in which the cosine function is positive.Also, 12 is an even integer. cos-1125° = cos1125° =cos90°×12 + 45° = cos 45° = 12

v We have:tan 7π4=tan 315°315° = 90°×3 + 45°315° lies in the fourth quadrant in which the tangent function  is negative.Also, 3 is an odd integer. tan 315° = tan 90°×3 + 45°  = -cot 45° = -1


vi We have:sin17π6=sin 510°510° = 90°×5 + 60°510° lies in the second quadrant in which the sine function is positive.Also, 5 is an odd integer. sin510° = sin 90°×5 + 60° =cos 60° = 12

vii We have:cos19π6=cos 570°570° = 90°×6 + 30° 570° lies in the third quadrant in which the cosine function is negative.Also, 6 is an odd integer. cos570° = cos 90°×6 + 30° =-cos 30° = -32

viii We have:sin-11π6=sin -330°sin -330° =-sin 330°= -sin 90°×3 + 60°330° lies in the fourth quadrant in which the sine function is negative.Also, 3 is an odd integer.sin-330° =-sin330° =-sin90°×3 + 60° = --cos60° =-- 12 =12

ix We have:cosec-20π3=cosec -1200°cosec -1200° =-cosec 1200°= -cosec 90°×13 + 30°1200° lies in the second quadrant in which the cosec function  is positive.Also, 13 is an odd integer. cosec-1200° = -cosec1200° =-cosec90°×13 + 30° = -sec30° = -23

x We have:tan -13π4=tan -585°tan -585° =-tan 585°= -tan 90°×6 + 45°585° lies in the third quadrant in which tangent function is positive.Also, 6 is an even integer. tan -585° = -tan 585° =-tan 90°×6 + 45° = -tan 45° = -1

xi We have:cos19π4=cos 855°855° = 90°×9 + 45°855° lies in the second quadrant in which the cosine function is negative.Also, 9 is an odd integer. cos855° = cos90°×9 + 45° =-sin45° = -12

xii We have:sin41π4=sin 1845°1845° = 90°×20 + 45°1845° lies in the first quadrant in which the sine function is positive.Also, 20 is an even integer. sin 1845° = sin 90°×20 + 45° =sin 45° = 12

xiii We have:cos39π4=cos 1755°1755° = 90°×19 + 45°1755° lies in the fourth quadrant in which the cosine function is positive.Also, 19 is an odd integer. cos1755° = cos90°×19 + 45° =sin 45° = 12

xiv We have:sin151π6=sin 4530°4530° = 90°×50 + 30°4530° lies in the third quadrant in which the sine function is negative.Also, 50 is an even integer. sin 4530° = sin 90°×50 + 30° =-sin 30° = -12

Page No 5.39:

Question 2:

Prove that:
(i) tan 225° cot 405° + tan 765° cot 675° = 0
(ii) sin8π3cos23π6+cos13π3sin35π6=12
(iii) cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = 12
(iv) tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
(v) cos 570° sin 510° + sin (−330°) cos (−390°) = 0
(vi) tan11π3-2sin4π6-34cosec2π4+4cos217π6=3-432
(vii) 3sinπ6secπ3-4sin5π6cotπ4=1

Answer:

i LHS = tan225°cot405° + tan765°cot675°                 =tan 90°×2+45°cot 90°×4+45° + tan 90°×8+45° cot 90°×7+45°                 =tan 45° cot 45° + tan 45°-tan 45°                 = 1×1 + 1×-1                 = 1-1                 = 0                 = RHSHence proved.

ii LHS = sin8π3cos23π6+cos13π3sin35π6                  = sin 83×180° cos 236×180°+cos 133×180° sin 356×180°                  =sin 480° cos 690°+cos 780° sin 1050°                  =sin 90°×5+30° cos 90°×7+60°+cos 90°×8+60° sin 90°×11+60°                  =cos 30° sin 60°+cos 60°-cos 60°                  =32×32+12×-12                  =34 - 14                  = 24                  =12                  = RHSHence proved.

iii LHS = cos 24° + cos 55° + cos 125° + cos 204° + cos 300°                   = cos 24° + cos 90°-35° + cos 90°×1+35°+ cos 90°×2+24° + cos 90°×3+30°                   =cos 24° + sin 35° - sin 35°- cos 24° + sin 30°                   =0 + 0+ 12                   = 12                   = RHSHence proved.

iv LHS = tan -225° cot -405° - tan -765° cot 675°                =- tan 225°-cot 405° - -tan 765° cot 675°                tan -x = tan x and cot -x = -cot x                  = tan 225° cot 405° +tan 765° cot 675°                  =tan 90°×2+45° cot 90°×4+45° + tan 90°×8+45° cot 90°×7+45°                  =tan 45° cot 45° + tan 45°-tan 45°                  = 1×1 + 1×-1                  = 1-1                  = 0                  = RHSHence, proved.

v LHS =cos 570°sin 510° + sin -330°cos -390°                  =cos 570° sin 510° + -sin 330°cos 390°                        sin-x = -sin x and cos-x = cos x                        =cos 570°sin510° -sin 330° cos 390°                 =cos 90°×6+30° sin 90°×5+60° -sin 90°×3+60° cos 90°×4+30°                 =-cos 30° cos 60° --cos 60° cos 30°                 =-cos 30° cos 60° +cos 30° sin 60°                 = 0                 = RHSHence proved.

vi LHS = tan11π3-2sin4π6-34cosec2π4+4cos217π6                  = tan11π3-2sin4π6-34cosecπ42+4cos17π62                 = tan113×180°-2sin46×180°-34cosec180°42+4cos17×180°62                 = tan 660°-2sin 120°-34cosec45°2+4cos 510°2                 =tan 660°-2sin 120°-34cosec45°2+4cos 510°2                 =tan 90°×7+30°-2sin 90°×1+30°-34cosec45°2+4cos90°×5+60°2                 =-cot 30°-2cos 30°-34cosec 45°2+4-sin60°2                 =-cot 30°-2cos 30°-34cosec45°2+4sin 60°2                  =-3-232-3422+4322                 =-3-3-32+3                 =3-432                 = RHSHence proved.

vii LHS = 3sinπ6secπ3-4sin5π6cotπ4                    = 3sin180°6sec180°3-4sin5×180°6cot180°4                    =3sin30°sec60°-4sin150°cot45°                    =3sin30°sec60°-4sin90°×1+60°cot45°                    =3sin 30°sec 60°-4cos 60° cot 45°                    =3×12×2 - 4×12×1                    =3 - 2                    =1                    = RHSHence proved.

Page No 5.39:

Question 3:

Prove that

(i) cos (2π+x) cosec (2π+x) tan (π/2+x)sec(π/2+x)cos x cot(π+x)=1

(ii)  cosec(90°+x)+cot(450°+x)cosec(90°-x)+tan(180°-x)+tan(180°+x)+sec(180°-x)tan(360°+x)-sec(-x)=2

(iii) sin(180°+x) cos(90°+x) tan(270°-x) cot(360°-x)sin(360°-x) cos(360°+x) cosec(-x) sin(270°+x)=1

(iv) 1+cot x-secπ2+x1+cot x+secπ2+x=2cot x

(v) tan (90°-x) sec(180°-x) sin(-x)sin(180°+x) cot(360°-x) cosec(90°-x)=1

Answer:

i LHS = cos2π+x cosec2π+x tanπ2+xsec π2+x cos x cot π+x             =cos x cosec x - cot x-cosec xcos x cot x                   =-cos x cosec x cot x-cosec x cos x cot x             = 1             =RHSHence proved.    

ii LHS =cosec 90°+x+cot 450°+xcosec 90°-x+tan 180°-x + tan 180°+x+sec 180°-xtan 360°+x-sec -x                  =cosec90°+x+cot450°+xcosec 90°-x+tan180°-x + tan 180°+x+sec 180°-xtan 360°+x-sec -x                  =cosec90°+x+cot 90°×5+xcosec90°-x+tan 90°×2-x + tan 90°×2+x+sec 90°×2-xtan90°×4+x-sec-x                  =sec x+cot 90°×5+xcosec90°-x+tan 90°×2-x + tan 90°×2+x+sec 90°×2-xtan 90°×4+x-sec -x                  =sec x-tan xsec x-tan x + tan x-sec xtan x-sec x                  =1 + 1                  =2                  =RHSHence proved.
iii LHS =sin 180°+ xcos90°+ x tan 270°- xcot 360°- xsin 360°- xcos360°+ xcosec- x sin 270°+ x                   =sin 90×2°+ xcos90°×1+ x tan90°×3- x cot90°×4- xsin90°×4- xcos90°×4+ x cosec - x sin 90°×3+ x                 =-sin  x -sin  x cot  x-cot  x-sin  x cos  x -cosec  x-cos  x                =sin2 x cot2  xsin  x cosec  x cos  x cos  x               =sin2 x ×cos2 xsin2xsin x ×1sin x×cos2x              =cos2  xcos2  x               =1              =RHSHence proved.

iv  LHS = 1+cot x-secπ2+x1+cot x+secπ2+x                   =  1+cot x--cosec x 1+cot x+-cosec x                       = 1+cot x+cosec x  1+cot x-cosec x                   = 1+cot x+cosec x  1+cot x-cosec x                   = 1+cotx+cosec x  1+cot x-cosec x                   =1+cot x2 - cosec x2               

                   =1+cot2 x+2cot x-cosec2 x                   =2 cot x                1+cot2x=cosec2x                   =RHS Hence proved.     


v  LHS =tan 90°-x sec 180°-x sin -xsin180°+xcot 360°-xcosec 90°-x                   =tan 90°×1-x sec 90°×2-xsin -xsin 90°×2+x cot 90°×4-xcosec 90°×1-x                  =cot x-sec x-sin x-sin x-cot x sec x                  =cot x sec x sin xsin x cot x sec x                  =cos xsin x×1cos x×sin xsin x×cos xsin x×1cos x                 =11                 =1                 =RHSHence proved.



Page No 5.40:

Question 4:

Prove that: sin2π18+sin2π9+sin27π18+sin24π9=2

Answer:

LHS = sin2π18 +sin2π9+sin27π18+sin24π9             = sin2π18 +sin22π18+sin27π18+sin28π18             = sin2π18 +sin22π18+sin27π18+sin28π18             = sin2π18 +sin22π18+sin2π2-2π18+sin2π2-π18             = sin2π18 +sin22π18+cos22π18+cos2π18             = sin2π18 +cos2π18+sin22π18+cos22π18             =1+1             =2             =RHSHence proved.

Page No 5.40:

Question 5:

Prove that:
sec3π2-xsecx-5π2+tan5π2+xtanx-3π2=-1.

Answer:

LHS =sec3π2-xsecx-5π2+tan5π2+xtanx-3π2        =sec3π2-xsec-5π2-x+tan5π2+xtan-3π2-x        =sec3π2-xsec5π2-x+tan5π2+x-tan3π2-x        =sec3π2-xsec5π2-x-tan5π2+xtan3π2-x        =secπ2×3-xsecπ2×5-x-tanπ2×5+xtanπ2×3-x        =-cosec xcosec x--cot xcot x         =-cosec2x+cot2 x        =-cosec2x-cot2 x        =-1        =RHSHence proved.

Page No 5.40:

Question 6:

In a ∆ABC, prove that:
(i) cos (A + B) + cos C = 0
(ii) cosA+B2=sinC2
(iii) tanA+B2=cotC2

Answer:

i In ABC:A + B + C = π A + B = π-CNow, LHS = cosA + B + cos C                  = cosπ - C + cos C                  = -cosC + cos C         cosπ - C = -cosC                    =0                  =RHSHence proved.

ii In ABC:A + B + C = πA + B = π-CA + B2 = π-C2A + B2 = π2-C2Now, LHS = cosA+B2                   = cosπ2-C2                   = sin C2             cosπ2-θ = sin θ                  =RHSHence proved.

iii In ABC:A + B + C = πA + B = π-CA + B2 = π-C2A + B2 = π2-C2Now, LHS = tanA+B2                   = tanπ2-C2                   = cotC2         tanπ2-θ = cot θ                   =RHSHence proved.

Page No 5.40:

Question 7:

In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that
cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0

Answer:

A, B, C and D are the angles of a cyclic quadrilateral. A + C = 180° and B + D = 180°A = 180 -C and B = 180 -DNow, LHS = cos180°-A + cos180°+B + cos180°+C -sin90°+D                 = -cosA + -cos B + -cosC -cosD                 = -cosA -cos B -cosC -cosD                 = -cos180°-C -cos180°-D -cosC -cosD                 = --cos C --cos D -cos C -cos D                 = cos C +cos D -cosC -cos D                 =0                 =RHSHence proved.

Page No 5.40:

Question 8:

Find x from the following equations:
(i) cosecπ2+θ+x cos θ cotπ2+θ=sinπ2+θ(ii) x cotπ2+θ+tanπ2+θsin θ+cosecπ2+θ=0

Answer:

90°=π2
i We have: cosec90°+θ +  x cos θ cot90°+θ = sin90°+θ sec θ + x cos θ -tan θ = cos θ sec θ - x cosθ tanθ = cos θ sec θ - x cosθ×sin θcos θ = cos θ sec θ - x sin θ = cos θ sec θ - cos θ = x sin θ 1cos θ - cosθ = x sin θ 1 - cos2 θcos θ = x sin θ sin2θcosθ = x sin θ sin2 θcos θ sin θ = x sin θcos θ = x tan θ = x  x =tan θ 


ii We have: x cot90°+θ +tan90°+θ sin θ +cosec90°+θ =0  x -tan θ +-cot θ sin θ  +sec θ  =0  - x tan θ -cot θ  sin θ  +sec θ  =0  - x×sin θcos θ  -cos θ sin θ × sin θ +1 cos θ=0  - x×sin θcos θ -cos θ  +1 cos θ =0  - x sin θ-cos2 θ+1cos θ =0 
-x sin θ-cos2θ+1=0-xsin θ+sin2θ=0xsin θ=sin2θx=sin2 θsin θx=sin θ

Page No 5.40:

Question 9:

Prove that:
(i) tan 4π-cos3π2-sin5π6cos2π3=14

(ii) sin13π3sin8π3+cos2π3sin5π6=12

(iii) sin 13π3sin2π3+cos4π3sin13π6=12

(iv) sin10π3cos13π6+cos8π3sin5π6=-1

(v) tan5π4cot9π4+tan17π4cot15π4=0

Answer:

i 4π=720°, 3π2=270°, 5π6=150°, 2π3=120° LHS = tan720° -cos270° - sin150° cos120°             = tan90°×8+0° -cos90°×3+0° - sin90°×1+60° cos90°×1+30°             = tan0° -sin0° - cos60° -sin30°             = tan0° -sin0° + cos60° sin30°             = 0-0 + 12× 12             = 14             =RHSHence proved.

ii 13π3=780°, 8π3=480°,2π3=120°,5π6=150° LHS = sin 780° sin 480° + cos 120° sin150°               =sin 90°×8+60° sin 90°×5+30° + cos 90°×1+30° sin 90°×1+60°               = sin 60° cos 30° + -sin 30° cos 60°               =sin 60° cos 30° - sin 30° cos60°                = 32×32-12×12              =34-14             = 12             =RHSHence proved.

iii 13π3=780°, 2π3=120°,4π3=240°,13π6=390°LHS = sin780° sin120° + cos240° sin390°               =sin90°×8+60° sin90°×1+30° + cos90°×2+60° sin90°×4+30°               = sin 60° cos 30° + -cos 60° sin 30°               =sin 60° cos 30°  -cos 60°sin 30°               = 32×32-12×12               =34-14               = 12               =RHSHence proved.

iv 10π3=600°, 13π6=390°,8π3=480°,5π6=150°LHS = sin 600°cos 390°+cos 480° sin 150°               =sin 90°×6+60° cos90°×4+30°+cos90°×5+30° sin90°×1+60°               = -sin 60° cos30°+ -sin 30° cos 60°               =-sin 60° cos30° -sin 30° cos 60°               = -32×32-12×12              =-34- 14              = -1              =RHSHence proved.

v 5π4=225°, 9π4=405°,17π4=765°,15π4=675°LHS = tan 225°cot 405°+tan 765° cot 675°               = tan90°×2+45°cot90°×4+45°+tan90°×8+45° cot90°×7+45°                           =tan 45°cot 45°+tan 45° -tan45°               =tan 45°cot 45°-tan 45° tan 45°               =1×1-1×1               =1-1               = 0               =RHSHence proved.

Page No 5.40:

Question 1:

Write the maximum and minimum values of cos (cos x).

Answer:

We know: -1  cos x 1Also, cos-θ = cosθWhen the angle increases from 0 to π2, the value of cosθ decreases.Maximum value of coscosx = cos0= 1'And, minimum value of coscosx = cos1 

Page No 5.40:

Question 2:

Write the maximum and minimum values of sin (sin x).

Answer:

We know:-1  sin x 1Also, sin-θ = -sinθWhen the angle increases from 0 to π2, the value of sinθ also increases.Maximum value of sinsinx = sin1And, minimum value of sinsinx = sin-1 = -sin1

Page No 5.40:

Question 3:

Write the maximum value of sin (cos x).

Answer:

We know:-1  cos x 1Also, sin-θ = -sinθWhen the angle increases from 0 to π2, the value of sinθ also increases. Maximum value of sincos x = sin1

Page No 5.40:

Question 4:

If sin x = cos2 x, then write the value of cos2 x (1 + cos2 x).

Answer:

We have: sinx = cos2x                        1 cos2x 1+cos2x=sinx1+sinx           Using 1=sinx +sin2x=sinx +1-cos2x=sinx +1-sinx          Using 1=1

Page No 5.40:

Question 5:

If sin x + cosec x = 2, then write the value of sinn x + cosecn x.

Answer:

We have:sinx + cosecx = 2sinx + 1sinx = 2sin2x +1sinx=2

sin2x+1 = 2 sinxsin2x+1-2sin x = 0sin x-12 =0sin x  -1 = 0sin x = 1And, cosec x = 1sin x  = 1 sinnx+cosecnx = 1n+1n  = 1+1 = 2


Page No 5.40:

Question 6:

If sin x + sin2 x = 1, then write the value of cos12 x + 3 cos10 x + 3 cos8 x + cos6 x.

Answer:

We have:sinx + sin2x = 1                     1sinx = 1 - sin2xsinx = cos2x                       2Now, taking cube of 1:sinx + sin2x = 1 sinx + sin2x 3= 13 sinx 3 + sin2x3  + 3sinx 2 sin2x+3sinx sin2x2=1 sinx 3 + sinx6  + 3sinx 4 +3sinx 5=1  sinx6 +3sinx 5 + 3sinx 4 + sinx 3=1  cos2x6 +3cos2x5 + 3cos2x 4 + cos2x 3=1  cos12x+3cos10x + 3cos8x + cos6x =1      

Page No 5.40:

Question 7:

If sin x + sin2 x = 1, then write the value of cos8 x + 2 cos6 x + cos4 x.

Answer:

We have:sinx + sin2x=1                     1sinx=1-sin2xsinx=cos2x                     2Now, taking square of 1: sinx+sin2x 2=12 sinx 2+sin2x2+2sinx  sin2x=1 sinx 2+sinx4+ 2sinx 3=1  sinx2+2sinx 3+sinx 4=1  cos2x2+2cos2x3+cos2x 4=1  cos4x+2cos6x+cos8x=1 cos8x+2cos6x+cos4x=1

Page No 5.40:

Question 8:

If sin θ1 + sin θ2 + sin θ3 = 3, then write the value of cos θ1 + cos θ2 + cos θ3.

Answer:

Sine function can take the maximum value of 1.
If, sinθ1+sinθ2+sinθ3=3, then we have:

 sinθ1 = 1

⇒ θ1=π2
Similarly, θ2=θ3=π2

cosθ1=cosθ2=cosθ3=0cosθ1+cosθ2+cosθ3=0

Page No 5.40:

Question 9:

Write the value of sin 10° + sin 20° + sin 30° + ... + sin 360°.

Answer:

sin10°+sin20°+...+sin170°+sin180°+sin360°-170°+sin360°-160°+...+sin360°-20°+sin360°-10°+sin360°=sin10°+sin20°+...+sin180°-sin170°-sin160°-...-sin20°-sin10°+sin360°            sin360°-x=- sinx=sin180°+sin360°=0+0=0

Page No 5.40:

Question 10:

A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm. Write the measure of the angle subtended by it at the centre of the loop.

Answer:

Circumference of the circle of radius 15 cm:
    2πr=2×3.14×15 cm= 94.2 cm

Now, 94.2 cm will be the length of arcl for the circle with radius 120 cm.
We know:
l=rθHere, θ is measured in radians. 94.2 = 120×θθ=0.785 radians

 45° = π4=227×4=0.785 radians
Therefore, the angle subtended by it at the centre of the loop is 45°.



Page No 5.41:

Question 11:

Write the value of 2 (sin6 x + cos6 x) −3 (sin4 x + cos4 x) + 1.

Answer:

2sin6x+cos6x-3sin4x+cos4x+1=2sin2x+cos2xsin4x+cos4x-sin2x.cos2x-3sin4x+cos4x+1=2.1sin4x+cos4x-sin2x.cos2x-3sin4x+cos4x+1=2sin4x+cos4x-2sin2x.cos2x-3sin4x+cos4x+1=-sin4x+cos4x-2sin2x.cos2x+1=-sin4x+cos4x+2sin2x.cos2x+1=-sin2x+cos2x2+1=-1+1=0

Page No 5.41:

Question 12:

Write the value of cos 1° + cos 2° + cos 3° + ... + cos 180°.

Answer:

cos1°+cos2°+cos3°+...+cos180°=cos1°+cos2°+cos3°+...+cos88°+cos89°+cos90°+cos180-89°+cos180-88°+...+cos180-1°+ cos180°      cos180°-θ=-cos θ=cos1°+cos2°+cos3°+...+cos88°+cos89°+cos90°-cos89°-cos88°-...-cos1°+ cos180°=cos90°+cos180°=0-1=-1

Page No 5.41:

Question 13:

If cot (α + β) = 0, then write the value of sin (α + 2β).

Answer:

cot α+β=0α+β=π2              1β=π2-α                   2α=π2-β                   3Now, sinα+2β = sinα+β+β                             =sinπ2+π2-α                             =sinπ-α                             =sin αNow, sinα+2β = sinα+2β                             =sinπ2-β+2β                             =sinπ2+β                             =cos β

Page No 5.41:

Question 14:

If tan A + cot A = 4, then write the value of tan4 A + cot4 A.

Answer:

tanA+cotA=4Squaring both the sides:tan2A+cot2A+2=16tan2A+cot2A=14Squaring both the sides again:tan4A+cot4A+2=196tan4A+cot4A=194

Page No 5.41:

Question 15:

Write the least value of cos2 x + sec2 x.

Answer:

We know:
cos x can take the minimum value of -1.

cos2 x + sec2 x

=cos4x+1cos2x=-14+1-12=2

Page No 5.41:

Question 16:

If x = sin14 x + cos20  x, then write the smallest interval in which the value of x lie.

Answer:

If x = 0°, 90°, 180°, 270°, 360°, then sin14x+cos20x will always be 1.

The smallest interval in which the value of x lie is (0,1].

Page No 5.41:

Question 17:

If 3 sin x + 5 cos x = 5, then write the value of 5 sin x − 3 cos x.

Answer:

 3 sin x +5 cos x=5      GivenSquaring both the sides: 9 sin2 x +25 cos2 x+30 sin x cos x=2530 sin x cos x=25-9 sin2 x -25 cos2 x               (1)We have to find the value of 5 sin θ-3 cos θ.5 sin x-3 cos x2=25 sin2 x +9 cos2 x -30  sin x cos x5 sin x-3 cos x2=25 sin2 x +9 cos2 x -25-9 sin2 x -25 cos2 x      From (1)5 sin x-3 cos x2=34 sin2x+34 cos2 x-255 sin x-3 cos x2=34-25         sin2x+cos2 x=15 sin x-3 cos x2=95 sin x-3 cos x=±3

Page No 5.41:

Question 1:

If tan x = x-14x, then sec x − tan x is equal to
(a) -2x,12x
(b) -12x,2x
(c) 2x
(d) 2x,12x

Answer:

(a) -2x,12x

We have,tanx = x - 14xsec2x = 1 + tan2x sec2x= 1 +x - 14x2sec2x  =x2 +116x2+12sec2x =x + 14x2secx =± x + 14xsecx - tanx= x + 14x - x - 14x  or  -x + 14x - x - 14x                           =12x or -2x

Page No 5.41:

Question 2:

If sec x=x+14x, then sec x + tan x =
(a) x,1x
(b) 2x,12x
(c) -2x,12x
(d) -1x,x

Answer:

(b) 2x,12x

We have,secx = x + 14xsec2x = =x2 +116x2+121+tan2x=1+x2 +116x2-12tan2x   = x2 +116x2-12tan2x=x - 14x2tanx =± x - 14xsecx - tanx = x + 14x - x - 14x  or  x + 14x -- x - 14x                           =12x or 2x

Page No 5.41:

Question 3:

If π2<x<3π2, then 1-sin x1+sin x is equal to
(a) sec x − tan x
(b) sec x + tan x
(c) tan x − sec x
(d) none of these

Answer:

(c) tan x − sec x

1-sin x1+sin x =1-sin x1-sin x1+sin x1-sin x=1-sin x21-sin2 x=1-sin x2cos2x=1-sin x-cos x          as,π2<x<3π2, so cosθ will be negative=-sec x -tan x =-sec x +tan x

Page No 5.41:

Question 4:

If π < x <2π, then 1+cos x1-cos x is equal to
(a) cosec x + cot x
(b) cosec x − cot x
(c) −cosec x + cot x
(d) −cosec x − cot x

Answer:

(d) −cosec x − cot x

1+cos x1-cos x =1+cos x1+cos x1-cos x1+cos x=1+cos x21-cos2x=1+cosx2sin2x=1+cosx-sinx          as,π<x<2π, so sin x will be negative=-cosecx +cotx =-cosecx -cotx

Page No 5.41:

Question 5:

If 0<x<π2, and if y+11-y=1+sin x1-sin x, then y is equal to
(a) cotx2

(b) tanx2

(c) cotx2+tanx2

(d) cotx2-tanx2

Answer:

(b) tanx2

We have:y+11-y=1+sin x1-sin x y+11-y=cos2x2+sin2x2+2sinx2cosx2cos2x2+sin2x2-2sinx2cosx2y+11-y=cosx2+sinx22cosx2-sinx22y+11-y=cosx2+sinx2cosx2-sinx2           0<x<π20<x2<π4, 0 to π4 cos x is greater than sin xy+11-y=cosx2cosx2+sinx2cosx2cosx2cosx2-sinx2cosx2  1+y1-y=1 + tanx21 - tanx2 Comparing both the sides:y = tanx2

Page No 5.41:

Question 6:

If π2<x<π, then 1-sin x1+sin x+1+sin x1-sin x is equal to
(a) 2 sec x
(b) −2 sec x
(c) sec x
(d) −sec x

Answer:

(b)  −2 sec x

1-sin x1+sin x +1+sin x1-sin x=1-sin x1-sin x1+sin x1-sin x+1+sin x1+sin x1-sin x1+sin x=1-sin x21-sin2x+1+sin x21-sin2x=1-sin x2cos2x+1+sin x2cos2x=1-sin x-cos x +1+sin x-cosx         π2<x<π, so cos x will be negative.=-sec x -tan x -sec x +tan x=-2sec x 

Page No 5.41:

Question 7:

If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then x2 + y2 + z2 is independent of
(a) θ, ϕ
(b) r, θ
(c) r, ϕ
(d) r.

Answer:

(a) θ, ϕ

We have:
x = r sin θ cos ϕ  ,  y = r sin θ sin ϕ and z = r cos θ,
x2 + y2 + z2
= r sinθ cosϕ2+r sinθ sinϕ2+ r cosθ2= r2 sin2θ cos2ϕ+r2 sin2θ sin2ϕ + r2 cos2θ = r2 sin2θ cos2ϕ+sin2ϕ + r2 cos2θ = r2 sin2θ×1 + r2 cos2θ= r2 sin2θ + r2 cos2θ= r2 sin2θ +cos2θ= r2 ×1=r2 Thus, x2 + y2 + z2 is independent of θ and ϕ.

Page No 5.41:

Question 8:

If tan x + sec x = 3, 0 < x < π, then x is equal to

(a) 5π6

(b) 2π3

(c) π6

(d) π3

Answer:

(c) π6

We have:tan x + sec x = 3            0 <x<π sec x + tan x= 31cos x+sin xcos x = 31 + sin x = 3cos x
1+sin x2  =3 cos x21+sin2x+2sin x = 3cos2x1+sin2x+2sin x= 3(1-sin2x)4 sin2x+2sin x =22 sin2x+sin x -1=0sinx = -1, 12Since 0<x<π, sin x cannot be negative. sin x=12 x=π6  

Page No 5.41:

Question 9:

If tan x=-15 and θ lies in the IV quadrant, then the value of cos x is
(a) 56

(b) 26

(c) 12

(d) 16

Answer:

(a) 56

In the fourth quadrant, cos x and sec x are positive.cosx = 1secx         =1sec2x         =11+tan2x        =1 1 + -152        =1 65        =56



Page No 5.42:

Question 10:

If 3π4<α<π, then 2cot α+1sin2α is equal to
(a)  1 − cot α
(b) 1 + cot α
(c) −1 + cot α
(d) −1 −cot α

Answer:

(d) −1 −cot α

We have: 2cotα+1sin2α    = 2cosαsinα+1sin2α= 2sin αcos α+1sin2α= 2sin αcosα+sin2α+cos2αsin2α= sinα + cosα2sin2α= 1+cot α2= 1+cot α=-1+cot α       When 3π4<α<π, cot α<-1cot α +1<0=-1-cot α

Page No 5.42:

Question 11:

sin6 A + cos6 A + 3 sin2 A cos2 A =
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(b) 1

We have:sin6A + cos6A + 3sin2A cos2A=sin2A3 + cos2A3 + 3sin2A cos2A×1=sin2A3 + cos2A3 + 3sin2A cos2Asin2A+cos2A=sin2A+cos2A3=13=1

Page No 5.42:

Question 12:

If cosec x-cot x=12, 0<x<π2,, then cos x is equal to
(a) 53

(b) 35

(c) -35

(d) -53

Answer:

(b) 35

We have: cosec x - cot x = 12                                 11cosec x - cot x = 2cosec2x - cot2xcosec x - cot x = 2cosec x + cot xcosec x - cot xcosec x- cot x = 2 cosec x + cot x= 2                             2Adding 1 and 2:2cosec x =12+22cosec x =52cosecx =541sinx =54sinx = 45
Now, 0<θ<π2 cosθ = 1-sin2θ       =1-452      =35

Page No 5.42:

Question 13:

If cosec x+cot x=112, then tan x =
(a) 2122

(b) 1516

(c) 44117

(d) 11744

Answer:

(c) 44117

We have: cosec x+ cotx=112                  11cosecx + cotx=211cosec2x - cot2xcosecx + cotx=211cosecx + cotxcosecx - cotxcosecx+ cotx = 211 cosecA - cotx= 211            2Subtracting 2 from 1:2cotx =112-2112cotx=121-4222cotx=11722cotx =117441tanx =11744tanx =44117

Page No 5.42:

Question 14:

sec2x=4xy(x+y)2 is true if and only if
(a) x + y ≠ 0
(b) x = y, x ≠ 0
(c) x = y
(d) x ≠0, y ≠ 0

Answer:

(b) x = y, x ≠ 0

We have:sec2x=4xy(x+y)24xy(x+y)21        sec2x14xy (x+y)2
4xy  x2+y2+2xy2xy  x2+y2x-y2 0x-y 0 x= yFor x = 0, sec2x will not be defined, x  0 x=y

Page No 5.42:

Question 15:

If x is an acute angle and tan x=17, then the value of cosec2 x-sec2 xcosec2 x+sec2 x is
(a) 3/4
(b) 1/2
(c) 2
(d) 5/4

Answer:

(a) 3/4

We have:tan x = 17 tan2 x =17Now, dividing the numerator and the denominator of cosec2 x-sec2 xcosec2 x+sec2 x by cosec2 x:
1-tan2x1+tan2x= 1-171+17= 68 = 34
 

Page No 5.42:

Question 16:

The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is
(a) 7
(b) 8
(c) 9.5
(d) 10

Answer:

(c) 9.5

We have:sin25° + sin210° + sin215° + ...+ sin285° + sin290°=sin25° + sin210° + sin215° + ...+ sin290° -10°+ sin290° -5°+ sin290°=sin25° + sin210° + sin215° + ...+cos210°+ cos25°+ sin290°=sin25°+ cos25° + sin210° +cos210°++ sin215° +cos215°+ sin220° +cos220°+ sin225° +cos225°+ sin230° +cos230° + sin235° +cos235°+ sin240° +cos240° +sin245° + sin290°=1+1+1+1+1+1+1+1+122+12         sin2θ+cos2θ = 1=8+12+1=9.5

Page No 5.42:

Question 17:

sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =
(a) 1
(b) 4
(c) 2
(d) 0

Answer:

(c) 2

We have: sin2π18 +sin2π9+sin27π18+sin24π9= sin2π18 +sin22π18+sin27π18+sin28π18= sin2π18 +sin22π18+sin27π18+sin28π18= sin2π18 +sin22π18+sin2π2-2π18+sin2π2-π18= sin2π18 +sin22π18+cos22π18+cos2π18= sin2π18 +cos2π18+sin22π18+cos22π18=1+1=2

Page No 5.42:

Question 18:

If tan A + cot A = 4, then tan4 A + cot4 A is equal to
(a) 110
(b) 191
(c) 80
(d) 194

Answer:

(d) 194

We have:tan A + cot A = 4Squaring both the sides:tan A + cot A2 = 42 tan2A + cot2A +2 tan Acot A = 16 tan2A + cot2A +2  = 16 tan2A + cot2A  = 14Squaring both the sides again: tan2A + cot2A2  = 142 tan4A + cot4A +2 tan2Acot2A = 196 tan4A + cot4A +2 = 196 tan4A + cot4A= 194

Page No 5.42:

Question 19:

If x sin 45° cos2 60° = tan260° cosec30°sec45° cot2° 30°, then x =
(a) 2
(b) 4
(c) 8
(d) 16

Answer:

(c) 8

We have:x sin 45° cos2 60° = tan2 60° cosec 30°sec45°cot230°x× 12 × 122 = 3 2 × 22 × 32x42 = 632  x = 632 ×42x=8

Page No 5.42:

Question 20:

If A lies in second quadrant 3tanA + 4 = 0, then the value of 2cotA − 5cosA + sinA is equal to

(a) -5310                           (b) 2310                           (c) 3710                           (d) 710                          

Answer:

It is given that π2<A<π.

3tanA+4=0tanA=-43cotA=-34

Now,

secA=±1+tan2A=±1+169=±259=±53secA=-53      A lies in 2nd quadrantcosA=-35

Also,

sinA=±1-cos2A=±1-925=±1625=±45sinA=45          A lies in 2nd quadrant

So,

2cotA-5cosA+sinA=2×-34-5×-35+45=-32+3+45=-15+30+810=2310

Hence, the correct answer is option B.

Page No 5.42:

Question 21:

If cosec x+cot x=112, then tan x =
(a) 2122

(b) 1516

(c) 44117

(d) 11743

Answer:

(c) 44117

We have: cosec x+ cot x = 112                11cosec x + cot x = 211cosec2 x - cot2xcosec x + cot x = 211cosec x + cot xcosec x - cot xcosec x + cot x = 211 cosec x - cot x= 211           2Subtracting 2 from (1):2cot x =112-2112cot x =121-4222cot x =11722cot x =117441tan x =11744tan x =44117

Page No 5.42:

Question 22:

If tan θ + sec θ =ex, then cos θ equals

(a) ex+e-x2
(b) 2ex+e-x
(c) ex-e-x2
(d) ex-e-xex+e-x

Answer:

(b) 2ex+e-x

We have: tan θ +sec θ = ex sec θ+ tan θ = ex                       11secθ + tanθ = 1exsec2 θ - tan2θsec θ + tan θ = 1exsec θ + tan θsec θ - tan θsec θ + tan θ = 1ex sec θ - tan θ= 1ex             2Adding 1 and 2:2sec θ =ex + 1ex2sec θ =ex2 + 1exsec θ =e2x + 12exsec θ =12×e2x + 1exsec θ =12×ex + e-x1cos θ =ex + e-x2cos θ =2ex + e-x

Page No 5.42:

Question 23:

If sec x + tan x = k, cos x =

(a) k2+12k

(b) 2kk2+1

(c) kk2+1

(d) kk2-1

Answer:

(b) 2kk2+1

We have:sec x + tan x = k                    11sec x + tan x = 1ksec2 x - tan2 xsec x + tan x = 1ksec x + tan xsec x - tan xsec x + tan x = 1k sec x - tan x= 1k           2Adding 1 and 2:2sec x =k + 1k2sec x =k2 + 1ksec x=k2 + 12k1cos x =k2 + 12kcos x =2kk2 + 1



Page No 5.43:

Question 24:

If fx=cos2x+sec2x, then

(a) f(x) < 1                             (b) f(x) = 1                             (c) 1 < f(x) < 2                              (d) f(x) ≥ 2        

Answer:

fx=cos2x+sec2x      =cos2x+sec2x-2cosxsecx+2cosxsecx      =secx-cosx2+2fx2  x                        secx-cosx20 x

Hence, the correct option is answer D.

Page No 5.43:

Question 25:

Which of the following is incorrect?

(a) sin x=-15                 (b) cos x = 1                (c) sec x=12                 (d) tan x = 20           

Answer:

(a) sin x=-15 is correct as -1sin x1

(b) cos x = 1 is correct as -1cos x1

(c) sec x=12 is not correct as sec x(-, -1] [1, )

(d) tan x = 20 is correct as tan x can take any real value.

Hence, the correct answer is option C.

Page No 5.43:

Question 26:

The value of cos1° cos2° cos3° ... cos179° is

(a) 12                         (b) 0                          (c) 1                          (d) -1                          

Answer:

cos1° cos2° cos3° ... cos179°=cos1° cos2° cos3°... cos90° ... cos179°=0                             cos90°=0

Hence, the correct answer is option B.
 

Page No 5.43:

Question 27:

The value of tan1° tan2° tan3° ... tan89° is

(a) 0                                  (b) 1                                  (c) 12                                  (d) not defined                               

Answer:

We know that, tan90°-θ=cotθ

So,

tan89°=tan90°-1°=cot1°tan88°=tan90°-2°=cot2°tan87°=tan90°-3°=cot3°.                .                .                ..                .                .                .tan46°=tan90°-44°=cot44°

 tan1° tan2° tan3° ... tan89°=tan1° tan2° tan3° ... tan44° tan45° tan46° ... tan87° tan88° tan89°=tan1° tan2° tan3° ... tan44° tan45° cot44° ... cot3° cot2° cot1°=tan1°cot1°tan2°cot2° tan3°cot3° ... tan44°cot44°tan45°=1              tan45°=1 and tanθcotθ=1

Hence, the correct answer is option B.

Page No 5.43:

Question 28:

Which of the following is correct?

(a) sin1°>sin1                    (b) sin1°<sin1                    (c) sin1°=sin1                    (d) sin1°=π180sin1                    

Answer:

We know that, 1 radian is approximately 57º.

Also, the value of sinx is always increasing for 0x90° ( or sinx is an increasing function for 0x90°).

Now,

1°<57°Or 1°<1 radiansin1°<sin1

Hence, the correct answer is option B.



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