Rd Sharma Xi (2018) Solutions for Class 12 Science Math Chapter 9 Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle are provided here with simple step-by-step explanations. These solutions for Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle are extremely popular among class 12 Science students for Math Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle 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 9 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 9.28:

Question 1:

Prove that:
1-cos 2x1+cos 2x=tan x

Answer:

LHS=1-cos 2x1+cos 2x      =2sin2x2cos2x   1-cos2x=2sin2x and 1+cos2x=2cos2x      =sinxcosx      =tanx=RHSHence proved.  

Page No 9.28:

Question 2:

Prove that:
sin 2x1-cos 2x=cot x

Answer:

LHS=sin2x1-cos2x      =2sinx×cosx2sin2x    sin2x ,1-cos2x=2sin2x       =2sinx×cosx2sinx×sinx      =cosxsinx      =cotx=RHSHence proved.

Page No 9.28:

Question 3:

Prove that:
sin 2x1+cos 2x=tan x

Answer:

LHS=sin2x1+cos2x      =2sin x×cos x1+2cos2x-1  sin2x=2sinx×cosx and cos2x=2cos2x-1      =2sinx×cosx2cosx×cosx      =sinxcosx      =tanx=RHSHence proved.

Page No 9.28:

Question 4:

Prove that:
2+2+2 cos 4x= 2 cosx

Answer:

LHS=2+2+2cos4x      =2+21+cos4x            =2+2×2cos22x         2cos22x=1+cos4x      =2+2cos2x

      =21+cos2x      =2.2cos2x                         2cos2x=1+cos2x      =2cosx=RHSHence proved.

Page No 9.28:

Question 5:

Prove that:
1-cos 2x+sin 2x1+cos 2x+sin 2x=tan x

Answer:

LHS=1-cos2x+sin2x1+cos2x+sin2x

       =2sin2x+sin2x2cos2x+sin2x     2sin2x=1-cos2x and 2cos2x=1+cos2x       =2sin2x+2sinx cosx2cos2x+2sinx cosx  sin2x=2sinx cosx       =2sinxsinx+cosx2cosxcosx+sinx        =tanθ=RHSHence proved.

Page No 9.28:

Question 6:

Prove that:
sin x + sin 2x1+cos x+cos 2x=tan x

Answer:

LHS=sinx+sin2x1+cosx+cos2x      =sinx+sin2xcosx+1+cos2x

      =sinx+2sinx cosxcosx+2cos2x  sin2x=2sinx cosx and 2cos2x=1+cos2x

      =sinx 1+2cosxcosx 1+2cosx     =tanx=RHSHence proved.

Page No 9.28:

Question 7:

Prove that:
cos 2 x1+sin 2 x=tan π4-x

Answer:

LHS=cos2x1+sin2x

      =cos2x-sin2xsin2x+cos2x+2sinx×cosx  cos2x=cos2x-sin2x and sin2x+cos2x=1      =cosx-sinxcosx+sinxcosx+sinx2         a2-b2=a+ba-b      =cosx-sinxcosx+sinx

On dividing the numerator and denominator by cos x, we get
      =1-tanx1+tanx      =tanπ4-x=RHSHence proved.

Page No 9.28:

Question 8:

Prove that:
cos x1-sin x=tan π4+x2

Answer:

  LHS=cos x1-sin x
          =cos2x2-sin2x2sin2x2+cos2x2-2sinx2×cosx2    cosx=cos2x2-sin2x2, sinx=2sinx2cosx2 and sin2x2+cos2x2=1          =cosx2-sinx2cosx2+sinx2cosx2-sinx22          =cosx2+sinx2cosx2-sinx2

On dividing the numerator and denominator by cosx2, we get

          =1+tanx21-tanx2          =tanπ4+x2=RHSHence proved.

Page No 9.28:

Question 9:

Prove that:

cos2π8+cos23π8+cos25π8+cos27π8=2

Answer:

LHS=cos2π8+cos23π8+cos25π8+cos27π8      =cos2π8+cos23π8+cos2π-3π8+cos2π-π8      =cos2π8+cos23π8+-cos3π82+-cosπ82

      =cos2π8+cos23π8+cos23π8+cos2π8      =2cos2π8+cos23π8      =2cos2π8+cos2π2-π8      =2cos2π8+sin2π8      =2=RHSHence proved.

Page No 9.28:

Question 10:

Prove that:

sin2π8+sin23π8+sin25π8+sin27π8=2

Answer:

LHS=sin2π8+sin23π8+sin25π8+sin27π8      =sin2π2-3π8+sin2π2-π8+sin25π8+sin27π8      =cos23π8+sin2π8+sin2π-3π8+sin2π-π8

      =cos23π8+sin2π8+sin23π8+cos2π8      =cos2π8+sin2π8+cos23π8+sin23π8      =1+1=2=RHSHence proved.

Page No 9.28:

Question 11:

Prove that:
cos α + cos β2+sin α + sin β2= 4cos2α-β2

Answer:

LHS=cosα+cosβ2+sinα+sinβ2      =cos2α+cos2β+2cosαcosβ +sin2α+sin2β+2sinαsinβ      =(cos2α + sin2α)+(cos2β + sin2β)+2cosαcosβ+sinαsinβ      =1 +1 + 2cos(α-β)      =21 + cos(α-β)      =22cos2α-β2      =4cos2α-β2=RHSHence proved.

Page No 9.28:

Question 12:

Prove that:
sin2 π8+x2-sin2 π8-x2=12 sin x

Answer:

LHS=sin2π8+x2-sin2π8-x2      =121-cos2π8+x2-121-cos2π8-x2      =12cosπ4-x-cosπ4+x

Using the identity cosC-cosD=-2sinC+D2sinC-D2, we get

      =12-2sinπ4-x+π4+x2 sinπ4-x-π4+x2       =-sinπ4sin-x      =sinπ4sinx  sin-x= -sinx      =12sinx=RHSHence proved.

Page No 9.28:

Question 13:

Prove that:
1+cos2 2x=2 cos4 x+sin4 x

Answer:

LHS=1+cos22x
Using the identity cos2x=cos2x-sin2x, we get

LHS=1+cos2x-sin2x2      =1+cos4x+sin4x-2cos2x sin2x

      =cos2x+sin2x2+cos4x+sin4x-2cos2xsin2x   cos2x+sin2x=1     =cos4x+sin4x+2cos2xsin2x+cos4x+sin4x-2cos2xsin2x     = 2(cos4x+sin4x)=RHSHence proved.

Page No 9.28:

Question 14:

Prove that:
cos3 2x+3 cos 2x=4cos6 x-sin6x

Answer:

RHS=4cos6x-sin6x       =4cos2x3-sin2x3

Using the identity  a3-b3=a-ba2+ab+b2, we get

       =4cos2x-sin2x cos4x+sin4x+sin2xcos2x        =4cos2x-sin2x cos4x+sin4x+sin2xcos2x

       =4cos2xcos2x-sin2x2+2sin2xcos2x+sin2xcos2x            =4cos2xcos22x+3sin2xcos2x       =4cos2xcos22x+31-cos2x21+cos2x2
       =4cos2xcos22x+341-cos22x       =cos2x4cos22x+31-cos22x       =cos2x4cos22x+3-3cos22x       =cos2xcos22x+3       =cos32x+3cos2x=LHSHence proved.

Page No 9.28:

Question 15:

Prove that:
sin 3x+sin x sin x+cos 3x-cos x cos x=0

Answer:

LHS=sin3x+sinx sinx+cos3x-cosxcosx

Using the identities sinC+sinD=2sinC+D2cosC-D2 and cosC-cosD=-2sinC+D2sinC-D2, we get

LHS=2sin3x+x2×cos3x-x2×sinx+-2sin3x+x2×sin3x-x2cosx       =2sin2x×cosx×sinx-2sin2x×sinx cosx       =0=RHSHence proved.

Page No 9.28:

Question 16:

Prove that:

cos2 π4-x-sin2 π4-x=sin 2x

Answer:

LHS=cos2π4-x-sin2π4-x       =cos2π4-x           cos2α-sin2α=cos2α       =cosπ2-2x       =sin2x=RHS            cosπ2-2α=sin2αHence proved.

Page No 9.28:

Question 17:

Prove that:
cos 4x=1-8 cos2x+8 cos4 x

Answer:

LHS=cos4x      =cos2×2x      =2cos2×2x-1   cos2θ=2cos2θ-1      =22cos2x-12-1 cos22θ=2cos2θ-12

      =24cos4x-4cos2x+1-1      =8cos4x-8cos2x+1      =1-8cos2x+8cos4x=RHSHence proved.

Page No 9.28:

Question 18:

Prove that:
sin 4x=4 sin x cos3x-4 cos x sin3 x

Answer:

LHS=sin 4x      =2sin2x cos2x   sin2θ=2sinθcosθ

Now, using the identities sin2α=2sinαcosα and cos2α=cos2α-sin2α, we get

LHS=2(2sinx cosx).(cos2x-sin2x)      =4sinx cos3x-4sin3x cosx=RHSHence proved.

Page No 9.28:

Question 19:

Show that: 3 sin x-cos x4+6 sin x+cos2+4 sin6 x+cos6 x=13

Answer:

LHS=3sinx-cosx4+6sinx+cosx2+4sin6x+cos6x      =3sinx-cosx22+6sin2x+cos2x+2sinxcosx+4sin2x3+cos2x3

      =3(sin2x+cos2x-2sinxcosx)2+6(1+sin2x)+4(sin2x+cos2x)(sin4x-sin2xcos2x+cos4x)      =3(1-sin2x)2+6+6sin2x+4(sin4x-sin2xcos2x+cos4x)      =31-sin2x2+6+6sin2x+4(sin2x)2+(cos2x)2-12sin2xcos2x

      =31-sin2x2+6+6sin2x+4sin2x+cos2x2-2sin2xcos2x-12sin2xcos2x      =3(1-sin2x)2+6+6sin2x+41-2sin2xcos2x-12sin2xcos2x      =31-2sin2x+sin22x+6+6sin2x+4-20sin2xcos2x      =-6sin2x+3sin22x+13+6sin2x-52sinxcosx2      =3sin22x+13-5sin22x      =13-2sin22x=RHSHence proved. 

Page No 9.28:

Question 20:

Show that: 2 sin6 x+cos6x-3 sin4 x+cos4 x+1=0

Answer:

LHS=2sin6x+cos6x−3sin4x+cos4x+1      =2sin2x3+cos23-3sin4x+cos4x+1

      =2sin2x+cos2xsin4x-sin2xcos2x+cos4x-3sin4x+cos4x+1      =2sin4x+cos4x-2sin2xcos2x-3sin4x+cos4x+1      =-sin4x+cos4x+2sin2xcos2x+1      =-1+1=0=RHSHence proved.

Page No 9.28:

Question 21:

Prove that: cos6 A-sin6 A=cos 2A1-14sin2 2A

Answer:

LHS=cos6A-sin6A      =cos2A3-sin2A3      =cos2A-sin2Acos4A+sin2A.cos2A+sin4A

      =cos2Acos4A+2sin2Acos2A+sin4A-sin2Acos2A      =cos2Asin2A+cos2A2-14×4sin2Acos2A

      =cos2Asin2A+cos2A2-142sinAcosA2      =cos2A1-14sin2A2      =cos2A1-14sin22A=RHSHence proved.

Page No 9.28:

Question 22:

Prove that: tanπ4+x+tanπ4-x=2 sec 2x

Answer:

LHS=tanπ4+x+tanπ4-x      =tanπ4+tanx1-tanπ4tanx+tanπ4-tanx1+tanπ4tanx    tanA+B=tanA+tanB1-tanAtanB and tanA-B=tanA-tanB1+tanAtanB

      =1+tanx1-tanx+1-tanx1+tanx      =1+tanx2+1-tanx21+tanx1-tanx      =2(1+tan2x)1-tan2x=2sec2x1-sin2xcos2x

      =2sec2xcos2xcos2x   cos2x-sin2x=cos2x      =2×1cos2x      =2sec2x=RHSHence proved.

Page No 9.28:

Question 23:

Prove that: cot2 x-tan2 x=4 cot 2x cosec 2x

Answer:

LHS=cot2x-tan2x      =cos2xsin2x-sin2xcos2x      =cos2x2-sin2x2sin2xcos2x

      =cos2x+sin2xcos2x-sin2xsin2xcos2x      =1×cos2xsin2xcos2x       cos2x-sin2x=cos2x      =4cos2x4sin2xcos2x

      =4cos2xsin2x2      =4cos2xsin2x×1sin2x      =4cot2x cosec2x=RHSHence proved.

Page No 9.28:

Question 24:

Prove that:
cos 4x-cos 4α=8 cos x-cos α cos x+cos α cos x-sin α cos x+sin α

Answer:

RHS=8cosx-cosα cosx+cosα cosx-sinα cosx+sinα      =8cos2x-cos2α cos2x-sin2α      =8cos4x-cos2x×sin2α-cos2α×cos2x+cos2α×sin2α      =8cos4x-cos2xsin2α+cos2α+cos2α×sin2α      =8cos4x-cos2x+cos2α×1-cos2α      =8cos4x-cos2x+cos2α-cos4α      =8cos2xcos2x-1+cos2α×1-cos2α

      =812cos2x2cos2x-2+12cos2α×2-2cos2α      =812cos2x2cos2x-1-1-12cos2α×2cos2α-1-1      =812cos2xcos2x-1-12cos2α×cos2α-1     cos2α=2cos2α-1       =8142cos2xcos2x-1-2cos2α×cos2α-1      =8141+cos2xcos2x-1-1+cos2αcos2α-1

      =814cos22x-1-cos22α+1      =8182cos22x-2cos22α      =1+cos4x-1+cos4α       = 1+cos4x-1-cos4α      =cos4x-cos4α=LHSHence proved.

Page No 9.28:

Question 25:

Prove that sin 3x+sin 2x-sin x=4 sin x cosx2 cos3x2

Answer:

LHS= sin3x+sin2x-sinx        =sin3x+2sin2x-x2cos2x+x2           sinA-sinB=2sinA-B2cosA+B2        =sin3x+2sinx2cos3x2        =2sin3x2cos3x2+2sin3x2cosx2        sin2A=2sinAcosA       =2cos3x2sin3x2cosx2       =2cos3x22sin3x2+x22cos3x2-x22      sinA+sinB=2sinA+B2cosA-B2       =2cos3x22sinxcosx2      =4sinxcosx2cos3x2=RHSHence proved.



Page No 9.29:

Question 26:

tan 821°2=3+2 2+1=2+3+4+6

Answer:

Here,tan82.5°=tan90-7.5°                  =cot7.5°                  =1tan7.5°We know,tanx2=sinx1+cosxOn putting x=15°, we get tan152°=sin15°1+cos15°                 =sin45-30°1+cos45-30°                 =sin45°cos30°-sin30°cos45°1+cos45°cos30°+sin45°sin30°                 =12×32-12×121+12×32+12×12                =322-1221+322+122                =3-122+3+1Now,  tan82.5°=1tan7.5°                  =22+3+13-1                  =22+3+13-1×3+13+1                  =3+122+3+132-12                  =26+3+3+22+3+13-1                  =26+23+22+42                  =6+3+2+2                  =2+3+4+6            ...1                  =6+3+2+2                  =32+1+22+1                  =3+22+1               ...2From eqs. 1 and 2, we get tan82.5°=3+22+1=2+3+4+6                

Page No 9.29:

Question 27:

Prove that: cot π8=2+1

Answer:

π8=2212°Let A=2212°Using the identity cot2A=cot2A-12cotA, we getcot45°=cot22212°-12cot2212°    1=cot22212°-12cot2212°         cot45°=12cot2212°-cot22212°+1=0  

cot22212-2cot2212-1=0cot22212-2cot2212+1-2=0cot2212-12=2cot2212-1=2cot2212=2+1

Page No 9.29:

Question 28:

(i) If cos x=-35 and x lies in the IIIrd quadrant, find the values of
cosx2, sinx2, sin 2x.

(ii) If  cos x=-35 and x lies in IInd quadrant, find the values of sin 2x and sinx2

Answer:

(i) cos x=-35

Using the identity cos2θ=cos2θ-sin2θ, we get

cosx=cos2x2-sin2x2-35=2cos2x2-11-35=2cos2x225=2cos2x215=cos2x2cosx2=±15

It is given that x lies in the third quadrant. This means that x2 lies in the second quadrant.

cosx2=-15
Again,

cosx=cos2x2-sin2x2-35=-152-sin2x2-35=15-sin2x2-15-35=-sin2x245=sin2x2sinx2=±25

It is given that x lies in the third quadrant. This means that x2 lies in the second quadrant.

sinx2=25

Now,sinx = 1-cos2xsinx=1--352sinx=1-925=±45

Since x lies in the third quadrant, sinx is negative.

sinx = -45sin2x=2sinxcosxsin2x=2×-45×-35sin2x=2425

(ii)  cos x=-35
sinx=1-cos2x=1--35sinx=±45
Here, x lies in the second quadrant.

sinx=45

We know,

sin2x = 2sinx cosx

sin2x=2×45×-35=-2425
Now,
cosx=1-2sin2x22sin2x2=1--35=85sin2x2=45sinx2=±25

Since x lies in the second quadrant, x2 lies in the first quadrant.
sinx2=25

Page No 9.29:

Question 29:

If sin x=53 and x lies in IInd quadrant, find the values of cosx2, sinx2 and tan x2

Answer:

Given:
sinx=53
Using the identity cosx=1-sin2x, we get

cosx=1-sin2x=1-532=±23

Since x lies in the 2nd quadrant, cosx is negative.

cosx=-23

Now,

cosx=1-2sin2x2-23=1-2sin2x2sinx2=±56

Since x lies in the 2nd quadrant, x2 lies in the 1st quadrant.

sinx2=56

Again,

cosx=2cos2x2-1-23=2cos2x2-1cosx2=±16cosx2=16   x2<π2
Now, tanx2=sinx2cosx2=5616=5

Page No 9.29:

Question 30:

(i) If 0 ≤ xπ and x lies in the IInd quadrant such that sin x=14. Find the values of cosx2, sinx2 and tanx2
(ii) If cos x=45 and x is acute, find tan 2x
(iii) If sin x=45 and 0<x<π2, find the value of sin 4x.

Answer:

(i) sin x=14

sinx=1-cos2x142=1-cos2x116-1=-cos2x1516=cos2xcosx=±154

Since x lies in the 2nd quadrant, cos x is negative.

Thus,

cosx=-154

Now, using the identity cosx=2cos2x2-1, we get

-154=2cos2x2-1-158=cos2x2-12cos2x2=4-158cosx2=±4-158

Since x lies in the 2nd quadrant and x2 lies in the 1st quadrant, cosx2 is positive.

cosx2=4-158

Again,

cosx=cos2x2-sin2x2-154=4-1582-sin2x2-154=4-158-sin2x2sin2x2=4+158sinx2=±4+158=4+158

Now,

tanx2=sinx2cosx2        =4+1584-158=4+154-15        =4+154+154-154+15        =4+1542-152=4+1516-15=4+15

(ii) cos x=45

sinx=1-cos2x           =1-452           =1-1625           =25-1625           =925           =35

 tan x=sin xcos x              =3545              =34

Now,
tan 2x=2 tanx1-tan2x           =2341-342           =2341-916           =32716           =247

Hence, the value of tan 2x is 247.

(iii) sin x=45 and 0<x<π2.

sin x=1-cos2 x452=1-cos2 x1625-1=-cos2 x925=cos2 xcos x=±35

Since x lies in the 1st quadrant, cos x is positive.

Thus,

cos x=35

Now,
sin 4x=2 sin 2x cos2x             =22 sin x cos x1-2 sin2 x             =22×45×351-2452             =224251-3225             =2242525-3225             =22425-725             =-336625

Hence, the value of sin 4x is -336625.

Page No 9.29:

Question 31:

If tanx=ba, then find the value of a+ba-b+a-ba+b.                            [NCERT]

Answer:

Given: tanx=ba

a+ba-b+a-ba+b=1+ba1-ba+1-ba1+ba=1+tanx1-tanx+1-tanx1+tanx=1+sinxcosx1-sinxcosx+1-sinxcosx1+sinxcosx
=cosx+sinxcosx-sinx+cosx-sinxcosx+sinx=cosx+sinx+cosx-sinxcosx-sinxcosx+sinx=2cosxcos2x-sin2x=2cosxcos2x

Page No 9.29:

Question 32:

If tan A=17 and tan B=13, show that cos 2A = sin 4B

Answer:

Given:
tan A=17 and tan B=13

Using the identity tan2B=2tanB1-tan2B, we get

tan2B=2×131-19=34
Now, using the identities cos2A=1-tan2A1+tan2A and sin4B=2tan2B1+tan22B, we get

cos2A=1-1721+172 and sin4B=2×341+342cos2A=4850 and sin4B=2×34×1625cos2A=2425 and sin4B=2425

∴ cos 2A = sin 4B

Page No 9.29:

Question 33:

Prove that:
cos 7° cos 14° cos 28° cos 56°=sin 68°16 cos 83°

Answer:

LHS=cos7° cos14° cos28° cos56°

On dividing and multiplying by 2sin7, we get

       =12sin7°×2sin7°×cos7°×cos14°×cos28°×cos56°       =2sin14°2×2sin7°×cos14°×cos28°×cos56°         =2sin28°2×4sin7°×cos28°×cos56°

       =2sin56°2×8sin7°×cos56°       =sin112°16sin7°       =sin180°-68°16sin90°-83°       =sin68°16cos83°       sin180°-θ=sinθ & sin90°-θ=cosθ       =RHSHence proved.

Page No 9.29:

Question 34:

Prove that:
 cos2π15 cos4π15 cos 8π15 cos 16π15 =116

Answer:

LHS=cos2π15 cos4π15 cos8π15  cos16π15 

On dividing and multiplying by 2sin2π15, we get

     =12sin2π15×2sin2π15×cos2π15 ×cos4π15 ×cos8π15 × cos16π15     =12×2sin2π15×2sin4π15×cos4π15 ×cos8π15 × cos16π15        =12×4sin2π152sin8π15×cos8π15 × cos16π15     =12×8sin2π152sin16π15 × cos16π15     =116sin2π15sin32π15 

     =-116sin2π15sin2π-32π15      sin2π-θ=-sinθ     =-116sin2π15sin-2π15     =116=RHSHence proved.

Page No 9.29:

Question 35:

Prove that: cosπ5cos2π5cos4π5cos8π5=-116

Answer:

cosπ5cos2π5cos4π5cos8π5=12sinπ52sinπ5cosπ5cos2π5cos4π5cos8π5            Multiplying and dividing by 12sinπ5=12sinπ5sin2π5cos2π5cos4π5cos8π5             sin2A=2sinAcosA=14sinπ52sin2π5cos2π5cos4π5cos8π5                       Multiplying and dividing by 2                  
=14sinπ5sin4π5cos4π5cos8π5=18sinπ52sin4π5cos4π5cos8π5               Multiplying and dividing by 2=18sinπ5sin8π5cos8π5=116sinπ52sin8π5cos8π5                            Multiplying and dividing by 2 
=sin16π516sinπ5=sin3π+π516sinπ5=-sinπ516sinπ5                       sin3π+θ=-sinθ=-116

Page No 9.29:

Question 36:

Prove that:
cos π65 cos 2π65 cos4π65 cos8π65 cos16π65 cos32π65=164

Answer:

LHS=cosπ65 cos2π65cos4π65 cos8π65cos16π65 cos32π65

On dividing and multiplying by 2sinπ65, we get

       =12sinπ65×2sinπ65×cosπ65 ×cos2π65×cos4π65× cos8π65×cos16π65× cos32π65       =2×sin2π652×2sinπ65×cos2π65×cos4π65× cos8π65×cos16π65× cos32π65        =2×sin4π652×4sinπ65×cos4π65× cos8π65×cos16π65× cos32π65       =2×sin8π652×8sinπ65× cos8π65×cos16π65× cos32π65

       =2×sin16π652×16sinπ65×cos16π65× cos32π65       =2×sin32π652×32sinπ65× cos32π65       =sin64π6564sinπ65=sinπ-π6564sinπ65       =sinπ6564sinπ65   sinπ-θ=sinθ       =164=RHSHence proved.

Page No 9.29:

Question 37:

If 2 tan α=3 tan β, prove that tan α-β=sin 2β5-cos 2β

Answer:

Given:
2 tan α=3 tan β

LHS=tanα-tanβ1+tanα×tanβ       =32×tanβ-tanβ1+32tan2β  2tanα=3tanβ       =12×tanβ1+32tan2β=tanβ2+3tan2β       =sinβcosβ2+3sin2βcos2β=sinβcosβ×cos2β2cos2β+3sin2β

       =sinβ×cosβ2cos2β+2sin2β+sin2β       =122sinβ×cosβ2cos2β+sin2β+sin2β       =12sin2β2+sin2β=sin2β4+2sin2β       =sin2β4+21-cos2β=sin2β6-2cos2β       =sin2β6-1+cos2β   1+cos2β=2cos2β       =sin2β5-cos2β=RHSHence proved.

Page No 9.29:

Question 38:

If sin α + sin β=a and cos α+cos β=b, prove that

(i) sin α+β=2aba2+b2

(ii) cos α-β=a2+b2-22

Answer:

The given equations are sin α + sin β=a and cos α+cos β=b.

(i)

 sinC+sinD=2sinC+D2cosC-D22sinα+β2cosα-β2=a          ...(1) 

Now, using the identity sinC+sinD=2sinC+D2cosC-D2 for the LHS of cos α+cos β=b, we get

2cosα+β2cosα-β2=b             ...(2) 

On dividing (1) by (2), we get

 tanα+β2=ab  

We know,

sinθ=2tanθ21+tan2θ2
sinα+β=2tanα+β21+tan2α+β2sinα+β=2×ab1+a2b2=2aba2+b2

(ii)

On squaring sinα+sinβ=a and cosα+cosβ=b and adding them, we get sin2α+sin2β+2×sinαsinβ+ cos2α+cos2β+2×cosαcosβ=a2+b21+1+2sinαsinβ+cosαcosβ=a2+b22sinαsinβ+cosαcosβ=a2+b2-22cosα-β=a2+b2-2       cosA-B=sinAsinB+cosAcosBcosα-β=a2+b2-22

Page No 9.29:

Question 39:

If 2 tanα2=tanβ2, prove that cos α=3+5 cos β5+3 cos β

Answer:

RHS=3+5cos β5+3cos β       =3+51-tan2 β21+tan2 β25+31-tan2 β21+tan2 β2       =3+3tan2 β2+5-5tan2 β25+5tan2 β2+3-3tan β2       =8-2tan2 β28+2tan2 β2       =8-8tan2 α28+8tan2 α2                              2tan α2=tan β2       =81-tan2 α281+tan2 α2       =1-tan2 α21+tan2 α2       =cos α =LHSHence proved.

Page No 9.29:

Question 40:

If cos x=cos α+cos β1+cos α cos β, prove that tanx2=±tanα2tanβ2

Answer:

Given:

cos x=cos α+cos β1+cos α cos β      ...(1)

1-tan2x21+tan2x2   =cosα+cosβ1+cosα×cosβ  cosx=1-tan2x21+tan2x2 By componendo and dividendo, we get1-tan2x2+1+tan2x21-tan2x2-1+tan2x2 =1+cosα×cosβ+cosα+cosβ-1+cosα cosβ-cosα-cosβ22tan2x2=1+cosα1+cosβ1-cosα1-cosβ

tan2x2=1-cosα1-cosβ1+cosα1+cosβtan2x2=2sin2α2×2sin2β22cos2α2×2cos2β2tan2x2=tan2α2×tan2β2tanx2=±tanα2×tanβ2Hence proved.

Page No 9.29:

Question 41:

If sec x+α+sec x-α=2 sec x, prove that cos x=±2 cosα2

Answer:

Equation sec x+α+sec x-α=2 sec x can be written as

1cosx+α+1cosx-α=2cosx1cosx×cosα-sinx×sinα+1cosx×cosα+sinx×sinα=2cosx     cosA+B=cosA×cosB-sinA×sinB and cosA-B=cosA×cosB+sinA×sinB 2cosx×cosαcos2x×cos2α-sin2x×sin2α=2cosxcosx×cosαcos2x×cos2α-1-cos2x×sin2α=1cosx

cos2x×cosαcos2x×cos2α-1-cos2x×sin2α=1cos2x×cosαcos2x×cos2α-sin2α+cos2xsin2α=1cos2x×cosα=cos2x×cos2α-sin2α+cos2xsin2αcos2x×cosα=cos2xcos2α+sin2α-sin2αcos2x×cosα=cos2x-sin2α

cos2x×cosα-cos2x=-sin2αcos2xcosα-1=-sin2αcos2x1-cosα=sin2αcos2x=sin2α2sin2α2       2sin2x2=1-cosx

cos2x=4sin2α2×cos2α22sin2α2      sin2x=4sin2x2×cos2x2  cosx=±2  cosα2Hence proved.



Page No 9.30:

Question 42:

If cos α + cos β=13 and sin sinα+sin β=14, prove that cosα-β2=±524

Answer:

Squaring and adding equations cos α + cos β=13 and sinα+sin β=14, we get

cos2α+cos2β+2cosα×cosβ+sin2α+sin2β+2sinα×sinβ=19+1161+1+2cosα×cosβ+sinα×sinβ=251442+2cosα-β=25144      cosA-B=cosA×cosB+sinA×sinBcosα-β=-263288            ... (1)

Now,

cos2α-β2=1+cosα-β2                  =1-2632882      [From (1)]                  =25576                  =±524

Page No 9.30:

Question 43:

If sin α=45 and cos β=513, prove that cosα-β2=865

Answer:

Given:
sin α=45 cos β=513.
Now,
cosα=1-sin2α=1-452=35And,sinβ=1-cos2α=1-5132=1213

Now,

cosα-β=cosα×cosβ+sinα×sinβcosα-β=35×513×45×1213=6365Thus,cosα-β2=1+cosα-β2              =1+63652              =865

Page No 9.30:

Question 44:

If a cos2x+b sin2x=c has α and β as its roots, then prove that

(i) tanα+tanβ=2ba+c             [NCERT EXEMPLAR]

(ii) tanα tanβ=c-ac+a

(iii) tanα+β=ba                  [NCERT EXEMPLAR]

Answer:

Given: a cos2x+b sin2x=c

a1-tan2x1+tan2x+b2tanx1+tan2x-c=0a1-tan2x+2b tanx-c1+tan2x=0a-a tan2x+2b tanx-c-c tan2x=0a+c tan2x-2b tanx+c-a=0                .....1

This a quadratic equation in terms of tan2x.

It is given that α and β are the roots of the given equation, so tanα and tanβ are the roots of (1).

Since tanα and tanβ are the roots of the equation a+c tan2x-2b tanx+c-a=0. Therefore,

(i)
tanα+tanβ=--2ba+c           Sum of roots=-batanα+tanβ=2ba+c

(ii)

tanα tanβ=c-aa+c           Product of roots=caOr tanα tanβ=c-ac+a

(iii)
tanα+β=tanα+tanβ1-tanα tanβ                 =2ba+c1-c-ac+a              From i and ii                 =2ba+cc+a-c+ac+a                 =2b2a                 =ba

Page No 9.30:

Question 45:

If cosα+cosβ=0=sinα+sinβ, then prove that cos2α+cos2β=-2cosα+β.                     [NCERT EXEMPLAR]

Answer:

Given: cosα+cosβ=0=sinα+sinβ

cosα+cosβ=0

Squaring on both sides, we get

cos2α+cos2β+2cosα cosβ=0                .....1

Also,

sinα+sinβ=0

Squaring on both sides, we get

sin2α+sin2β+2sinα sinβ=0                    .....2

Subtracting (2) from (1), we get

cos2α+cos2β+2cosα cosβ-sin2α+sin2β+2sinα sinβ=0cos2α-sin2α+cos2β-sin2β+2cosα cosβ -sinα sinβ=0cos2α+cos2β+2cosα+β=0cos2α+cos2β=-2cosα+β                   



Page No 9.36:

Question 1:

Prove that: sin 5x=5 sin x-20 sin3 x+16 sin5 x

Answer:

LHS=sin5x      =sin3x+2x      =sin3x×cos2x+cos3x×sin2x      =3sinx-4sin3x1-2sin2x+4cos3x-3cosx×2sinx cosx      =3sinx-6sin3x-4sin3x+8sin5x+8cos4x-6cos2xsinx      =3sinx-10sin3x+8sin5x+8sinx1-sin2x2-6sinx1-sin2x      =3sinx-10sin3x+8sin5x+8sinx1-2sin2x+sin4x-6sinx+6sin3x      =3sinx-10sin3x+8sin5x+8sinx-16sin3x+8sin5x-6sinx+6sin3x      =5sinx-20sin3x+16sin5x      =RHSHence proved.

Page No 9.36:

Question 2:

Prove that: 4 cos3 10°+sin3 20°=3 cos 10° + sin 20°

Answer:

We know,  sin60°=cos30°              =32sin3×20°=cos3×10°3sin20°-4sin320°=4cos310°-3cos10°                sin3θ=3sinθ-4sin3θ and cos3θ=4cos3θ-3cosθ 4cos310+sin320°=3cos10°+sin20°Hence proved.

Page No 9.36:

Question 3:

Prove that: cos3 x sin 3x+sin3 x cos 3x=34 sin 4x

Answer:

We know,cos3x=4cos3x-3cosxcos3x=cos3x+3cosx4 ...iAlso,sin3x=3sinx-4sin3xsin3x=3sinx-sin3x4 ...ii
Now,LHS=cos3x sin3x+sin3x cos3x       =cos3x+3cosx4sin3x+3sinx-sin3x4cos3x                                                                        Using i and ii      =143sin3x cosx+sinx cos3x+cos3x sin3x-sin3x cos3x      =143sin3x+x+0      =34sin4x      =RHSHence proved.

Page No 9.36:

Question 4:

tan x tanx+π3+tan x tanπ3-x+tanx+π3tanx-π3=-3

Answer:

π3=60°LHS=tanx tanx+60°+tanx tanx-60°+ tanx+60°tanx-60°      =tanxtanx+31-3tanx+tanxtanx-31+3tanx+tanx+31-3tanxtanx-31+3tanx                    tanA+B=tanA+tanB1-tanAtanB and tanA-B=tanA-tanB1+tanAtanB      =1+3tanxtanxtanx+3+1-3tanxtanxtanx-3+tan2x-31+3tanx1-3tanx      =tanx+3tan2xtanx+3+tanx-3tanxtanx-3+tan2x-31-3tan2x

Page No 9.36:

Question 5:

tan x+tanπ3+x-tanπ3-x=3 tan 3x

Answer:

π3=60°LHS    = tanx+tan60°+x-tan60°-x          =tanx+tan60°+tanx1-tan60° tanx-tan60°-tanx1+tan60° tanx                   tanx+y=tanx+tany1-tanx tany and tanx-y=tanx-tany1+tanx tany          =tanx+3+tanx1-3 tanx-3-tanx1+3 tanx          =tanx+3+3tanx+tanx+3 tan2x+3+3tanx+tanx-3tan2x1-3 tanx1+3 tanx          =tanx+8tanx1-3tan2x

       =tanx-3tan3x+8tanx1-3tan2x=9tanx-3tan3x1-3tan2x=33tanx-tan3x1-3tan2x        tan3θ=3tanθ-tan3θ1-3tan2θ =3tan3x=RHSHence proved.

Page No 9.36:

Question 6:

cot x+cotπ3+x+cotπ3-x=3 cot 3x

Answer:

π3=60°LHS=cotx+cot60°+x-cot60°-x      =1tanx+1tan60°+x-1tan60°-x

     =1tanx+1-3tanx3+tanx-1+3tanx3-tanx                       tanx+y=tanx+tany1-tanx tany and tanx-y=tanx-tany1+tanx tany=1tanx-8tanx3-tan2x=3-tan2x-8tan2xtanx3-tan2x=3-9tan2x3tanx-tan3x=31-3tan2x3tanx-tan3x=3×1tan3x     tan3θ=1-3tan2θ3tanθ-tan3θ=3cot3x=RHSHence proved.

Page No 9.36:

Question 7:

cot x+cotπ3+x+cot2π3+x=3 cot 3x

Answer:

π3=60°, 2π3=120°LHS=cotx+cot60°+x+cot120°+x      =cotx+cot60°+x-cot180°-120°+x                                       -cotθ=cot180°-θ      =cotx+cot60°+x-cot60°-x      =1tanx+1tan60°+x-1tan60°-x

     =1tanx+1-3tanx3+tanx-1+3tanx3-tanx                tanx+y=tanx+tany1-tanxtany and tanx-y=tanx-tany1+tanx tany=1tanx-8tanx3-tan2x=3-tan2x-8tan2x3tanx-tan3x=3-9tan2x3tanx-tan3x=31-3tan2x3tanx-tan3x=3×1tan3x        tan3θ=3tanθ-tan3θ1-3tan2θ=3cot 3x=RHSHence proved.

Page No 9.36:

Question 8:

sin 5x=5 cos4 x sin x-10 cos2x sin3 x+sin5 x

Answer:

LHS=sin5x      =sin3x+2x      =sin3x×cos2x+cos3x×sin2x      =3sinx-4sin3x2cos2x-1+4cos3x-3cosx×2sinx cosx      =-3sinx+4sin3x+6sinx cos2x-8sin3x cos2x+8sinx cos4x-6sinx cos2x      =8sinx cos4x-8sin3x cos2x-3sinx+4sin3x      =5sinx cos4x-10sin3x cos2x-3sinx+3sinx cos4x+4sin3x+2sin3x cos2x      =5sinx cos4x-10sin3x cos2x-3sinx1-cos4x+2sin3x2+cos2x

      =5sinxcos4x-10sin3xcos2x-3sinx1-cos2x1+cos2x+2sin3x2+cos2x      =5sinx cos4x-10sin3x cos2x-3sin3x1+cos2x+2sin3x2+cos2x      =5sinx cos4x-10sin3x cos2x-sin3x31+cos2x-22+cos2x      =5sinx cos4x-10sin3x cos2x-sin3x3+3cos2x-4-2cos2x      =5sinx cos4x-10sin3x cos2x-sin3x cos2x-1      =5sinx cos4x-10sin3x cos2x-sin3x×-sin2x      =5sinx cos4x-10sin3x cos2x+sin5x      =5cos4x sinx-10cos2x sin3x+sin5x      =RHSHence proved.



Page No 9.37:

Question 9:

sin3 x+sin32π3+x+ sin34π3+x=-34 sin 3x

Answer:

LHS=sin3x+sin32π3+x+sin34π3+x     =3sinx-sin3x4+3sin2π3+x-sin32π3+x4+3sin4π3+x-sin34π3+x4                                                                                                     sin3θ=3sinθ-sin3θ4      =3sinx-sin3x4+3sinπ-2π3+x-sin2π+3x4+3sinπ+π3+x-sin4π+3x4     =143sinx-sin3x+3sinπ3-x-sin3x-3sinπ3+x+sin3x     =143sinx-sin3x+3sinπ3-x-3sinπ3+x-sin3x-sin3x

    =143sinx-3sin3x+3sinπ3-x-sinπ3+x=143sinx-3sin3x+32cosπ3-x+π3+x2sinπ3-x-π3-x2                sinC-sinD=2cosC+D2sinC-D2=143sinx-3sin3x+6cosπ3sin-x=143sinx-3sin3x-3sinx=-34sinx=RHSHence proved.

Page No 9.37:

Question 10:

Prove that sin x sin π3-x sin π3+x14 for all values of x

Answer:

π3=60°We have,          sinx sin60-x sin60+x      =sinxsin260-sin2x            sinA+B sinA-B=sin2A-sin2B            =sinx34-sin2x              =14sinx3-4sin2x      =143sinx-4sin3x            =14sin3x                 3sinx-4sin3x=sin3x      14         sinx1 for all x sinx sin60-x sin60+x14Hence proved.

Page No 9.37:

Question 11:

Prove that cos x cos π3-x cos π3+x14 for all values of x

Answer:

π3=60°We have,cosx cos60°-x cos60°+x=cosxcos260°-sin2x     cos2A-sin2B=cosA-B cosA+B=cosx14-sin2x   =cosx141-4sin2x=14cosx1-41-cos2x=14cosx-3+4cos2x

=144cos3x-3cosx=14cos3x        cos3x=4cos3x-3cosx           14   cosx1 for all xcosx cos60°-x cos60°+x14



Page No 9.42:

Question 1:

Prove that:
sin22π5-sin2-π3=5-18

Answer:

2π5=72°, π3=60°LHS=sin272°-sin260°      =sin290°-18°-34      =cos218°-34          sin90°-θ=cosθ      =10+2542-34        cos18°=10+254      =10+2516-34

     =10+25-1216=25-216=5-18=RHSHence proved.

Page No 9.42:

Question 2:

Prove that:
sin2 24°-sin2 6°=5-18

Answer:

LHS=sin224°-sin26°      =sin24°+6° sin24°-6°   sinA+B sinA-B=sin2A-sin2B       =sin30° sin18°      =12×5-14           sin18°=5-14      =5-18      =RHSHence proved.

Page No 9.42:

Question 3:

Prove that:
sin2 42°-cos2 78=5+18

Answer:

LHS=sin242°-cos278°      =sin290°-48°-cos290°-12°      =cos248°-sin212°          =cos48°+12° cos48°-12°     cosA+B cosA-B=cos2A-sin2       =cos60° cos36°      =12×5+14         cos36°=5+14      =5+18      =RHSHence proved.

Page No 9.42:

Question 4:

Prove that:
cos 78° cos 42° cos 36°=18

Answer:

LHS=cos78° cos42° cos36°      =2cos78° cos42°2 cos36°      =cos78°+42°+cos78°-42°2 ×cos36°                        2cosAcosB=cosA+B+cosA-B      =12cos120°+cos36° cos36°

     =12-cos180°-120°+cos36° cos36°=12-cos60°+cos36° cos36°=12-12+5+145+14=12×5-14×5+14=18=RHSHence proved.

Page No 9.42:

Question 5:

Prove that:
cosπ15cos2π15cos4π15cos7π15=116

Answer:

LHS=cosπ15cos2π15cos4π15cos7π15

     =2sinπ15cosπ152sinπ15cos2π15cos4π15cos7π15                          On dividing and multiplying by 2sinπ15

    =2sin2π15×cos2π152×2sinπ15cos4π15cos7π15=2sin4π15×cos4π152×2×2sinπ15cos7π15=sin8π152×2×2sinπ15cos7π15

   =2sin8π15cos7π152×2×2×2sinπ15=2sin8π15cos7π1516sinπ15=sin8π15+7π15+sin8π15-7π1516sinπ15    2sinAcosB=sinA+B+sinA-B

   =sinπ+sinπ1516sinπ15=0+sinπ1516sinπ15=sinπ1516sinπ15=116=RHS
Hence proved.

Page No 9.42:

Question 6:

Prove that:
cos 6° cos 42° cos 66° cos 78°=116

Answer:

LHS=cos6° cos42° cos66° cos78°=142cos6° cos66°2cos42° cos78°     =14cos72°+cos60°cos120°+cos36°       2cosAcosB=cosA+B+cosA-B  =14cos90°-72°+12-12+5+14
=14sin18°+12-12+5+14=145-14+125+14-12=145-1+245+1-24=1645+15-1=1645-1=116=RHSHence proved.

Page No 9.42:

Question 7:

Prove that:
sin 6° sin 42° sin 66° sin 78°=116

Answer:

LHS=142 sin6° sin66°2sin42° sin78°       =14cos60°-cos72° cos36°-cos120°                                    2sinAsinB=cosA-B-cosA+B      =1412-sin18° 5+14+12

    =1412-5-145+14+12=142-5+14 5+1+24=1643-5 3+5=1649-5=116=RHSHence proved.

Page No 9.42:

Question 8:

Prove that:
cos 36° cos 42° cos 60° cos 78°=116

Answer:

LHS=cos36° cos42° cos60° cos78°      =12cos36° cos60° 2cos42° cos78°                        2cosAcosB=cosA+B+cosA-B      =125+14 ×12cos120°+cos36°            =5+116-12+5+14      =5+1 5-164      =5-164      =116      =RHSHence proved.

Page No 9.42:

Question 9:

Prove that :
sinπ5sin2π5sin3π5sin4π5=516

Answer:

LHS=sinπ5sin2π5sin3π5sin4π5        =122 sinπ5 sin4π5122 sin2π5 sin3π5        =14cosπ5-4π5-cosπ5+4π5cos2π5-3π5-cos2π5+3π5        =14cos-3π5-cos5π5cos-π5-cos5π5        =14cos3π5-cosπcosπ5-cosπ        =14cos3π5+1cosπ5+1        =14cosπ-2π5+1cosπ5+1        =14-cos2π5+15+14+1                cos π5=5+14        =14-5-14+15+14+1                cos 2π5=5-14        =14-5-145+14-5-14+5+14+1        =14-52-116+5+1-5+14+1        =14-416+24+1        =14-14+24+1        =14-1+2+44        =516        =RHS

Thus, LHS = RHS

Hence, sinπ5sin2π5sin3π5sin4π5=516.

Page No 9.42:

Question 10:

Prove that:
cosπ15 cos 2π15 cos 3π15 cos 4π15 cos 5π15 cos6π15 cos 7π15=1128

Answer:

LHS=cosπ15 cos2π15 cos4π15 cos3π15 cos5π15 cos6π15 cos7π15      =cosπ15 cos2π15 cos4π15cos3π15 cos6π15 ×-cos8π15      =-12cosπ15 cos2π15 cos4π15 cos8π15× 12× cos3π15 cos6π15      =-12×2324sinπ152sinπ15cosπ15  cos2π15 cos4π15  cos8π15 ×222×sin3π15 2sin3π15cos3π15 cos6π15      =-23132sinπ15sin2π15  cos2π15 cos4π15  cos8π15  ×24sin3π15 sin6π15 cos6π15      =-2232sinπ152sin2π15  cos2π15 cos4π15  cos8π15  ×14sin3π15 2sin6π15 cos6π15
     =-232sinπ15sin8π15 cos8π15 ×sin12π154sin3π15=-132sinπ15sin16π15  ×sin12π154sin3π15=-sinπ+π15128sinπ15×sinπ-3π15sin3π15=--sinπ15128sinπ15×sin3π15sin3π15=1128=RHSHence proved.

Page No 9.42:

Question 1:

If cos 4x=1+k sin2 x cos2 x, then write the value of k.

Answer:

We have,   cos4x=1+k sin2x cos2x   cos2×2x=1+k sin2x cos2x 1-2sin22x=1+k sin2x cos2x  1-22sinxcosx2=1+k sin2x cos2x  1-8sin2xcos2x=1+k sin2x cos2x  sin2xcos2xk+8=0k+8=0 k=-8

Page No 9.42:

Question 2:

If tanx2=mn, then write the value of m sin x + n cos x.

Answer:

Given:

tanx2=mn
sinx2cosx2=mnLet sinx2 be mk and cosx2 be nk.  Now, msinx+ncosx=2m sinx2cosx2+ncos2x2-sin2x2                         =2m×mk×nk+nn2k2-m2k2  

                         =2m2k2n+nk2n2-m2                         =nk22m2+n2-m2                         =nk2m2+n2                         =nm2k2+n2k2                         =nsin2x2+ cos2x2                         =n1msinx+ncosx=n

Page No 9.42:

Question 3:

If π2<x<3π2, then write the value of 1+cos 2x2.

Answer:

π2<x<3π21+cos2x2=2cos2x2=cosxIn second quadrant cosx is negative.1+cos2x2=-cosx 

Page No 9.42:

Question 4:

If π2<x < π, the write the value of 2+2+2 cos 2x in the simplest form.

Answer:

We have,

2+2+2cos2x= 2+21+cos2x                             =2+2.2cos2x                            =2+2cosx                            =2-2cosx   π2<x<π                              =21-cosx                            =2.2sin2x2                            =2sinx2                            =2sinx2    π4<x2<π2

Page No 9.42:

Question 5:

If π2<x< π, then write the value of 1-cos 2x1+cos 2x.

Answer:

We have,
1-cos2x1+cos2x=2sin2x2cos2x=sinxcosx=sinx-cosx     π2<x<π=-tanx

Page No 9.42:

Question 6:

If π<x<3π2, then write the value of 1-cos 2x1+ cos 2x.

Answer:

We have,1-cos2x1+cos2x=2sin2x2cos2x                   =sinxcosx                   =sinxcosx                    =-sinx-cosx     π<x<3π2  1-cos2x1+cos2x=tanx

Page No 9.42:

Question 7:

In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.

Answer:

Let, B=90°A+C=90°=π2C=π2-Asin C=sin π2-Asin C=cos A ...iNow,sin2 A+sin2 B+sin2 C= sin2 A+1+sin2 C        sin B=sin 90°=1                                 =sin2 A+cos2 A+1    Using i                                 =1+1                                 =2

Page No 9.42:

Question 8:

Write the value of cos2 76° + cos2 16° - cos 76° cos 16°.

Answer:

We have,cos276°+cos216°-cos76°cos16°=121+cos276°+1+cos216°-cos76+16°-cos76-16°             2cos2θ=1+cos2θ and 2cosA cos B=cosA+B +cosA-B=122+cos152°+cos32°-cos92°-12=1232-cos180-152°+cos32°-cos92°

=1232-cos28°+2sin92°+32°2 sin92°-32°2                         cosC-cosD=2sinC+D2sinD-C2=1232-cos28°+2sin124°2 sin60°2=1232-cos28°+2sin62° sin30°=1232-cos28°+2sin62°×12=1232-cos28°+sin62°=1232-cos28°+sin90-28°=1232-cos28°+cos28°=34

Page No 9.42:

Question 9:

If π4<x<π2, then write the value of 1-sin 2x.

Answer:

We have,1-sin2x=sin2x+cos2x-2sinx cosx=sinx-cosx2    =sinx-cosx=sinx-cosx                 sinx>cosx for π4<x<π21-sin2x=sinx-cosx

Page No 9.42:

Question 10:

Write the value of cosπ7 cos2π7 cos4π7.

Answer:

We have, cosπ7 cos2π7 cos4π7=2sinπ7 cosπ7 cos2π7 cos4π72sinπ7                                                                               On dividing and multiplying by 2sinπ7                                    =2×sin2π7  cos2π7 cos4π72×2sinπ7

Proceeding in the same way, we get

cosπ7 cos2π7 cos4π7=sin8π78sinπ7                                   =sinπ+π78sinπ7                                   =-sinπ78sinπ7cosπ7 cos2π7 cos4π7=-18

Page No 9.42:

Question 11:

If tanA=1-cosBsinB, then find the value of tan2A.

Answer:

Given:

tanA=1-cosBsinBtanA=2sin2B22sinB2cosB2                  1-cos2θ=2sin2θ and sin2θ=2sinθcosθtanA=sinB2cosB2=tanB2A=B2

2A=Btan2A=tanB 

Hence, the value of tan2A is tanB.

Page No 9.42:

Question 12:

If sinx+cosx=a, then find the value of sin6x+cos6x.

Answer:

Given: sinx+cosx=a

Squaring on both sides, we get

sin2x+cos2x+2sinxcosx=a21+2sinxcosx=a2sinxcosx=a2-12             .....1

Now,

sin6x+cos6x=sin2x+cos2x3-3sin2xcos2xsin2x+cos2x=1-3a2-122                         Using 1=4-3a2-124

Hence, the required value is 144-3a2-12.

Page No 9.42:

Question 13:

If sinx+cosx=a, find the value of sinx-cosx.

Answer:

Given: sinx+cosx=a

Now,

sinx+cosx2+sinx-cosx2=sin2x+cos2x+2sinxcosx+sin2x+cos2x-2sinxcosxsinx+cosx2+sinx-cosx2=2sin2x+cos2xsinx+cosx2+sinx-cosx2=2

a2+sinx-cosx2=2sinx-cosx2=2-a2sinx-cosx2=2-a2sinx-cosx=2-a2               x2=x

Thus, the required value is 2-a2.



Page No 9.43:

Question 1:

8 sinx8 cos x2cosx4 cosx8 is equal to

(a) 8 cos x
(b) cos x
(c) 8 sin x
(d) sin x

Answer:

(d) sin x

We have, 8sinx8 cosx2  cosx4  cosx8           =4×2sinx8 cosx8 cosx2  cosx4            =4×sinx4cosx2  cosx4           =2×2sinx4 cosx4 cosx2           =2×sinx2cosx2           =sinx

Page No 9.43:

Question 2:

sec 8A-1sec 4A-1=

(a) tan 2Atan 8A


(b) tan 8Atan 2A

(c) cot 8Acot 2A

(d) none of these.

Answer:

(b) tan 8Atan 2A

We have,sec8A-1sec4A-1=1cos8A-11cos4A-1                =cos4Acos8A×1-cos8A1-cos4A

            =cos4Acos8A×2sin24A2sin22A       2sin2θ=1-cos2θ=2cos4A sin4A sin4A2×cos8A sin22A=sin8A sin4Acos8A×2sin2A×sin2A=tan8A×2sin2A×cos2A2sin2A×sin2A=tan8A×cot2A=tan8Atan2A

Page No 9.43:

Question 3:

The value of cos π65 cos 2π65 cos 4π65 cos 8π65 cos 16π65 cos 32π65 is
(a) 18

(b) 116

(c) 132

(d) none of these

Answer:

(d) none of these

We have,cosπ65 cos2π65 cos4π65 cos8π65 cos16π65 cos32π65= 2sinπ652sinπ65 cosπ65 cos2π65 cos4π65 cos8π65 cos16π65 cos32π65                      dividing and multiplying by 2sinπ65= 2sin2π652×2sinπ65 cos2π65 cos4π65 cos8π65 cos16π65 cos32π65= 2sin4π652×4sinπ65 cos4π65 cos8π65 cos16π65 cos32π65
= 2sin8π652×8sinπ65 cos8π65 cos16π65 cos32π65= 2sin16π652×16sinπ65 cos16π65 cos32π65= 2sin32π652×32sinπ65 cos32π65= sin64π6564sinπ65= sinπ-π6564sinπ65=sinπ6564sinπ65=164

Page No 9.43:

Question 4:

If cos 2x+2 cos x=1 then, 2-cos2 x sin2 x is equal to
(a) 1

(b) -1

(c) -5

(d) 5

Answer:

(a) 1

We have,cos2x+2cosx=12cos2x-1+2cosx=1cos2x+cosx-1=0cosx=-1±12+42cosx=-1±52cosx=-1+52

Now,2-cos2xsin2x=2--1+522 1-cos2x                     =2-141-25+5 1-141-25+5                     =141+55-1=44=1

Page No 9.43:

Question 5:

For all real values of x, cot x-2 cot 2x is equal to
(a) tan 2x
(b) tan x
(c) -cot 3x
(d) none of these

Answer:

(b) tan x

We have,cotx-2cot 2x=cotx-2cot2x-12cotx                     =cot2x-cot2x+1cotx                     =1cotx                     =tanx

Page No 9.43:

Question 6:

The value of 2 tan π10+3 sec π10-4 cos π10 is
(a) 0
(b) 5
(c) 1
(d) none of these

Answer:

(a) 0

We have,2tanπ10+3secπ10-4cosπ10=2tan18°+3sec18°-4cos18°=2sin18°cos18°+3×1cos18°-4cos18°=2×5-1410+254+3×110+254-4×10+254=2×5-110+25+3×410+25-10+25=25-2+12-10+25210+25=25+10-10-2510+25=0

Page No 9.43:

Question 7:

If in a  ABC, tan A + tan B + tan C = 0, then cot A cot B cot C =
(a) 6
(b) 1
(c) 16
(d) none of these

Answer:

(d) none of these

ABC is a triangle.

A+B+C=πA+B=π-CtanA+B=tanπ-CtanA+tanB1-tanA tanB=-tanCtanA+tanB=-tanC+tanA tanB tanCtanA+tanB+tanC =tanA tanB tanC0=tanA tanB tanC    [Given: tanA tanB tanC=0]tanA tanB tanC=01tanA tanB tanC=10cotA cotB cotC 

Page No 9.43:

Question 8:

If cos x=12 a+1a, and cos 3 x = λ a3+1a3, then λ=
(a) 14

(b) 12

(c) 1

(d) none of these

Answer:

(b) 12
Given:cosx=12a+1a cos3x=λa3+1a3Now,cos3x=18a3+1a3+3a1aa+1acos3x=18a3+1a3+3×2cosx                    cosx=12a+1acos3x=18cos3xλ+6cosxcos3x=184cos3x-3cosxλ+6cosxcos3x=4cos3x8λ-3cosx8λ+6cosx8On comparing the powers of cos3x on both sides, we get1=48λλ=12

Page No 9.43:

Question 9:

If 2 tan α=3 tan β, then tan α - β=

(a) sin 2 β5-cos 2 β

(b) cos 2 β5-cos 2 β

(c) sin 2 β5+cos 2 β

(d) none of these

Answer:

(a) sin 2 β5-cos 2 β

Given: 2tanα=3tanβNow,tanα-β=tanα-tanβ1+tanαtanβ             =32tanβ-tanβ1+32tanβtanβ             =3tanβ-2tanβ2+3tan2β             =tanβ2+3tan2β             =sinβcosβ2+3sin2βcos2β             =sinβcosβ2cos2β+3sin2β             =sinβcosβ2+sin2β             =2sinβcosβ4+2sin2β             =sin2β4+1-cos2β tanα-β=sin2β5-cos2β

Page No 9.43:

Question 10:

If tan α=1-cos βsin β, then

(a) tan 3 α = tan 2 β

(b) tan 2 α = tan β

(c) tan 2 β=tan α

(d) none of these

Answer:

(b) tan 2 α = tan β

tanα=1-cosβsinβ      =2sin2β22sinβ2cosβ2      =sinβ2cosβ2tanα=tanβ2α=β22α=βtan2α=tanβ

Page No 9.43:

Question 11:

If sin α + sin β=a and cos α-cos β=b then tan α-β2=
(a) -ab

(b) -ba

(c) a2+b2

(d) none of these

Answer:

(b) -ba

Given: sinα+sinβ=a2sinα+β2cosα-β2=a      ...(1)Also,cosα+cosβ=b-2sinα+β2sinα-β2=b    ...(2)On dividing (1) by (2), we get-cosα-β2sinα-β2=ab-sinα-β2cosα-β2=batanα-β2=-ba

Page No 9.43:

Question 12:

The value of cot x2-tan x22 1-2 tan x cot 2 x is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(d) 4

We have,cotx2-tanx22 1-2tanx cot2xcot2x2-2cotx2tanx2+tan2x2 1-2tanx cot2x-12cotxcot2x2-2+tan2x21-tanx cot2x-1cotxcot2x2+tan2x2-21-cotx-tanxcotxcot2x2+tan2x2-2tan2xcot2x2+tan2x2-22tanx21-tan2x22
=11-tan2x224+4tan4x2-8tan2x2=11-tan2x22 4-8tan2x2+4tan4x2=41-tan2x22 tan2x22-2tan2x2+1=4tan2x2-121-tan2x22=4

Page No 9.43:

Question 13:

The value of tan x sin π2+x cos π2-x
(a) 1
(b) -1
(c) 12 sin 2x
(d) none of these.

Answer:

(d) none of these

We have,tanθ sinπ2+x cosπ2-x=sinxcosxcosx sinx=sin2x



Page No 9.44:

Question 14:

sin2 π18+ sin2π9+sin27π18+sin24π9=

(a) 1
(b) 2
(c) 4
(d) none of these.

Answer:

(b) 2

We have,sin2π18+sin2π9+sin27π18+sin24π9=121-cosπ9+1-cos2π9+1-cos7π9+1-cos8π9                          sin2θ=1-cos2θ2=124-cosπ9-cos2π9--cosπ-7π9--cosπ-8π9=124-cosπ9-cos2π9+cos2π9+cosπ9=42=2

Page No 9.44:

Question 15:

If 5 sin α=3 sin α+2 β 0, then tan α+β is equal to
(a) 2 tan β
(b) 3 tan β
(c) 4 tan β
(d) 6 tan β

Answer:

(c) 4 tan β

We have,5 sin α=3 sin α+2 β53=sin α+2 βsin α5-35+3=sin α+2 β-sin αsin α+2 β+sin α         Using componendo and dividendo28=sin α+2 β-sin αsin α+2 β+sin α14=2cosα+2 β+α2sinα+2 β-α22sinα+2 β+α2cosα+2 β-α214=cosα+ β sin βsinα+ β cos β14=cot α+ β tan β14=1 tan α+ βtan β tan α+ β=4 tan β

Page No 9.44:

Question 16:

The value of 2 cosx - cos 3x - cos 5x - 16 cos3x sin2x is
(a) 2
(b) 1
(c) 0
(d) −1

Answer:

(c) 0

We have,2cosx-cos3x-cos5x-16cos3x sin2x=2cosx-cos3x-cos5x-16cos3x+3cosx4×1-cos2x2=2cosx-cos3x-cos5x-2cos3x+3cosx1-cos2x=2cosx-cos3x-cos5x-2cos3x-cos3x cos2x+3cosx-3cosx cos2x=2cosx-cos3x-cos5x-2cos3x+3cosx+2cos3x cos2x+32cosx cos2x=2cosx-cos3x-cos5x-2cos3x+3cosx+cos5x+cosx+3cos3x+3cosx                                                               2cosAcosB=cosA+B+cosA-B=2cosx-cos3x-cos5x-2cos3x-6cosx+cos5x+cosx+3cos3x+3cosx=0

Page No 9.44:

Question 17:

If A=2 sin2x-cos 2x, then A lies in the interval
(a) -1, 3
(b) 1, 2
(c) -2, 4
(d) none of these

Answer:

(a) -1, 3

A=2sin2x-cos2x   =2sin2x-1-2sin2x   =4sin2x-10sin2x14×04×sin2x4×104sin2x40-14sin2x-14-1-14sin2x-13-1A3A-1,3

Page No 9.44:

Question 18:

The value of cos 3x2 cos 2x-1 is equal to
(a) cos x
(b) sin x
(c) tan x
(d) none of these

Answer:

(a) cos x

We have,cos3x2cos2x-1=4cos3x-3cosx22cos2x-1-1   cos3x=4cos3x-3cosx                          =4cos3x-3cosx4cos2x-2-1                          =4cos3x-3cosx4cos2x-3                          =cosx4cos2x-34cos2x-3                           =cosx

Page No 9.44:

Question 19:

If tan π/4+x+tan π/4-x=λ sec 2x, then
(a) 3
(b) 4
(c) 1
(d) 2

Answer:

(d) 2

Given: tanπ4+x+tanπ4-x=λ sec 2xtanπ4+tanx1-tanπ4×tanx+tanπ4-tanx1+tanπ4×tanx=λ sec 2x1+tanx1-tanx+1-tanx1+tanx=λ sec 2x1+tanx2+1-tanx21-tanx1+tanx=λ sec 2x21+tan2x1-tan2x=λ sec 2x

2sec2x1-tan2x=λ sec 2x2cos2x1-tan2x=λ sec 2x2cos2x1-sin2xcos2x=λ sec 2x2cos2x-sin2x=λ sec 2x2cos2x=λ sec 2x2sec2x=λ sec 2x2=λ λ=2

Page No 9.44:

Question 20:

The value of cos2π6+x- sin2π6-x is
(a) 12 cos 2x

(b) 0

(c) -12 cos 2x

(d) 12

Answer:

(a) 12 cos 2x

We have,cos2π6+x-sin2π6-x=cos2π6+x-cos2π2-π6-x=cos2π6+x-cos2π3+x=cosπ6+x+cosπ3+xcosπ6+x-cosπ3+x=2cosπ6+x+π3+x2 cosπ6+x-π3-x2 2sinπ6+x+π3+x2 sinπ3+x-π6-x2=4cosπ4+xcos-π12  sinπ4+x sinπ12=4cosπ4+xcosπ12  sinπ4+x sinπ12=2sinπ4+xcosπ4+x2 sinπ12cosπ12=sinπ2+2xsinπ6=cos2x×12=12cos2x

Page No 9.44:

Question 21:

sin 3x1+2 cos 2x is equal to

(a) cos x
(b) sin x
(c) – cos x
(d) sin x

Answer:

(b) sin x

We have,sin 3x1+2cos 2x=3sinx-4sin3x1+21-2sin2x                     =3sinx-4sin3x1+2-4sin2x                     =sinx3-4sin2x3-4sin2x                     =sin x

Page No 9.44:

Question 22:

The value of 2 sin2 B+4 cos A+B sin A sin B+cos 2 A+B is
(a) 0
(b) cos 3A
(c) cos 2A
(d) none of these

Answer:

(c) cos 2A

We have,2sin2B+4cosA+B sinA sinB+cos2A+B=1-cos2B+cos2A+B+4cosA+B sinA sinB=1+cos2A+B-cos2B+4cosA+B sinA sinB=1-2sinAsinA+2B+4cosA+B sinA sinB                          cosC-cosD=-2sinC+D2sinC-D2=1-2sinAsinA+2B-2sinBcosA+B=1-2sinAsinA+2B-sinB+A+B+sinB-A+B                            2sinCcosD=sinC+D+sinC-D=1-2sinAsinA+2B-sinA+2B+sin-A=1-2sinAsinA=1-2sin2A=cos2A

Page No 9.44:

Question 23:

The value of 2sin 2x + 2 cos2x-1cos x-sin x-cos 3x+sin 3x is
(a) cos x
(b) sec x
(c) cosec x
(d) sin x

Answer:

(c) cosec x

We have,2sin2x+2cos2x-1cosx-sinx-cos3x+sin3x=2sin2x+cos2xcosx-sinx-4cos3x+3cosx+3sinx-4sin3x=2sin2x+cos2x4cosx-4cos3x+2sinx-4sin3x=2sin2x+cos2x4cosx1-cos2x+2sinx1-2sin2x=2sin2x+cos2x4cosx sin2x+2sinx cos2x=2sin2x+cos2x2×2sinx cosx sinx+2sinx cos2x=2sin2x+cos2x2sin2x sinx+2sinx cos2x=2sin2x+cos2x2sinxsin2x+cos2x=1sinx=cosecx

Page No 9.44:

Question 24:

2 1-2 sin2 7x sin 3x is equal to
(a) sin 17x-sin 11x
(b) sin 11x-sin 17x
(c) cos 17x - cos 11x
(d) cos 17x + cos 11x

Answer:

(a) sin 17x-sin 11x

We have, 21-2sin2 7x sin 3x=2cos 14x sin3x                                              cos2x=1-2sin2x                                     =2 sin3x cos 14x                                     = sin 17x-sin 11x                                            2 sinA cosB=sinA+B-sinA-B  21-2sin2 7x sin 3x= sin 17x-sin 11x

Page No 9.44:

Question 25:

If α and β are acute angles satisfying cos 2 α=3 cos 2 β-13-cos 2 β, then tan α =
(a) 2 tan β

(b) 12tan β

(c) 2 cot β

(d) 12 cot β

Answer:

(a) 2 tan β

Given: cos2α=3cos 2β-13-cos2βcos2α-1cos2α+1=3cos 2β-1-3-cos2β3cos 2β-1+3-cos2β      Using componendo and dividendocos2α-1cos2α+1=4cos 2β-42cos 2β+2-1-cos2α1+cos2α=-41-cos 2β21+cos 2β1-cos2α1+cos2α=21-cos 2β1+cos 2β2sin2α2cos2α=22sin2β2cos2βtan2α=2tan2βtan α=2 tan β

Page No 9.44:

Question 26:

If tan x2=1-e1+e tan α2, then cos α=

(a) 1-e cos cos x+e

(b) 1+e cos xcos x-e

(c) 1-e cos xcos x-e

(d) cos x-e1-e cos x

Answer:

(d) cos x-e1-e cos x

Given: tanx2=1-e1+etanα2tanx2tanα2=1-e1+eSquaring both sides, we get,tan2x2tan2α2=1-e1+etan2α21-e=tan2x21+e

sin2α2cos2α21-e=sin2x2cos2x21+e121-cosα121+cosα1-e=121-cosx121+cosx1+e1-cosα1+cosx1-e=1+cosα1-cosx1+e1+cosx1-e-cosα1+cosx1-e=1-cosx1+e+cosα1-cosx1+ecosα1+cosx1-e+1-cosx1+e=1+cosx1-e-1-cosx1+ecosα=2cosx-2e2-2ecosx=cosx-e1-ecosx



Page No 9.45:

Question 27:

If 2n+1 x = π, then 2n cos x cos 2x cos 22x ... cos 2n-1 x=1
(a) -1
(b) 1
(c) 1/2
(d) None of these

Answer:

(b) 1

2n+1x=π      Given2nx+x=π2nx=π-xsin 2nx=sinπ-xsin 2nx =sin x            ...(1) 

2n cos x cos 2x cos 22x ... cos 2n-1 x=2n ×sin 2nx2nsin x                                                          =sin 2nxsin x                                                          =sin xsin x                   From (1)                                                          =1

Page No 9.45:

Question 28:

If tan x=t then tan 2x + sec 2x=

(a) 1+t1-t

(b) 1-t1+t

(c) 2t1-t

(d) 2t1+t

Answer:

(a) 1+t1-t

tan 2x + sec 2x=2 tan x1-tan2 x+1+tan2 x1-tan2 x                          =2tan x+1+tan2 x1-tan2x                          =1+tan x21-tan2 x                          =1+tan x1+tan x1+tan x1-tan x                          =1+tan x1-tan x                          =1+t1-t          tan x=t given

Page No 9.45:

Question 29:

The value of cos4 x+sin4 x-6 cos2 x sin2 x is
(a) cos 2x
(b) sin 2x
(c) cos 4x
(d) none of these

Answer:

(c) cos 4x

cos4 x+sin4x-6cos2x sin2x= cos4 x+sin4x-2cos2x sin2x-4cos2x sin2x                                         =cos2x-sin2x2-2sin x cos x2                                         =cos22x-sin22x                                         =cos4x

Page No 9.45:

Question 30:

The value of  cos 36°-A cos 36°+A+ cos 54°-A cos 54°+A is
(a) cos 2A
(b) sin 2A
(c) cos A
(d) 0

Answer:

(a) cos 2A

cos36°-Acos36°+A+cos 54°-Acos54°+A                =cos90°-54°+Acos90°-54°-A+cos 54°-Acos54°+A                =sin54°+Asin54°-A+cos54°-Acos54°+A           cos90°-θ=sinθ                =cos54°+A-54°+A                                                      cosA-B=cosAcosB+sinAsinB                  =cos 2A       

Page No 9.45:

Question 31:

The value of tan x tan π3-x tan π3+x is
(a) cot 3x
(b) 2cot 3x
(c) tan 3x
(d) 3 tan 3x

Answer:

(c) tan 3x

π3=60°tanx tan60°-x tan60°+x=tan x×tan60°-tanx1+tan60°tanx×tan60°+tanx1-tan60° tanx                                            =tan x×3-tanx1+3tanx×3+tanx1-3tanx                                            =tanx3-tan2x1-3tan2x                                            =3tanx-tan3x1-3tan2x                                            =tan 3x

Page No 9.45:

Question 32:

The value of tan x+tan π3+x+tan 2π3+x is
(a) 3 tan 3x
(b) tan 3x
(c) 3 cot 3x
(d) cot 3x

Answer:

(a) 3 tan 3x

π3=60°, 2π3=120°tan x+ tan60°+x+tan120°+x=tan x+tan60°+tanx1-tan60°tanx+tan 120°+tan x1-tan 120° tanx                                                     =tan x +3+tan x1-3tan x+-3+tanx1+3tanx                                                        =tan x1-3tan2x+3+tanx1+3tanx+-3+tanx1-3tanx1-3tan2x                                                     =tan x-3tan3x+3+3tanx+tanx+3tan2x+tanx-3tan2x-3+3tanx1-3tan2x                                                     =9tanx-3tan3x1-3tan2x                                                     =33tanx-tan3x1-3tan2x                                                     =3tan3x

Page No 9.45:

Question 33:

The value of sin 5 α-sin 3αcos 5 α+2 cos 4α+cos 3α=
(a) cot α/2
(b) cot α
(c) tan α/2
(d) None of these

Answer:

(c) tan α/2

sin5α-sin3αcos5α+2cos4α+cos3α= sin5α-sin3αcos5α+cos3α+2cos4α                                              =2sinαcos4α2cos4αcosα+2cos4α                                              =2sinαcos4α2cos4αcosα+1                                              =sinαcosα+1                                              =2sinα2cosα2cos2α2-sin2α2+sin2α2+cos2α2                                              =2sinα2cosα22cos2α2                                              =sinα2cosα2                                              =tanα2

Page No 9.45:

Question 34:

sin 5xsin x is equal to
(a) 16 cos4x-12 cos2x+1
(b) 16 cos4x+12 cos2x+1
(c) 16 cos4x-12 cos2x-1
(d) 16 cos4x+12 cos2x-1

Answer:

(a) 16 cos4x-12 cos2x+1

To find:sin 5xsinxNow,sin5x= sin3x+2x           = sin3xcos2x+cos3xsin2x           =3sinx-4sin3x1-2sin2x+4cos3x-3cosx2sinxcosx           =3sin x-6sin3x-4sin3x+8sin5x +2sinxcos2x4cos2x-3           =3sin x-10sin3x+8sin5x+2sinx1- sin2x41-sin2x-3           =3sin x-10sin3x+8sin5x+2sinx-2sin3x4-4sin2x-3           =3sin x-10sin3x+8sin5x+2sinx-8sin3x-2sin3x+8sin5x           =5sinx-20sin3x+16sin5xsin 5xsinx=5sinx-20sin3x+16sin5xsinx                    =5-20sin2x+16sin4x                    =5-201-cos2x+161-cos2x2                     =5-20+20cos2x+161+cos4x-2cos2x                  =5-20+20cos2x+16+16cos4x-32cos2x                  =16cos4x-12cos2x+1

Page No 9.45:

Question 35:

If n=1, 2, 3, ..., then cos α cos 2 α cos 4 α ... cos 2n-1 α is equal to

(a) sin 2n α2n sin α

(b)  sin 2n α2n sin 2n-1 α

(c) sin 4n-1 α4n-1 sin α

(d) sin 2n α2n sin α

(e) None of these

Answer:

(d) sin 2n α2n sin α

cos α cos 2 α cos 4α ... cos 2n-1 α=sin2nα2nsin α

Page No 9.45:

Question 36:

If tanx=ab, then b cos 2x+a sin 2x is equal to

(a) a                              

(b) b                              

(c) ab                              

(d) ba                              

Answer:

Given: tanx=ab
Now,

b cos2x+a sin2x=b 1-tan2x1+tan2x+ a2tanx1+tan2x=b1-a2b21+a2b2+a2×ab1+a2b2=bb2-a2a2+b2+2a2ba2+b2

=b3-a2b+2a2ba2+b2=b3+a2ba2+b2=bb2+a2a2+b2=b

Hence, the correct answer is option B.
Given: tanx=ab
Now,

b cos2x+a sin2x=b 1-tan2x1+tan2x+ a2tanx1+tan2x=b1-a2b21+a2b2+a2×ab1+a2b2=bb2-a2a2+b2+2a2ba2+b2

=b3-a2b+2a2ba2+b2=b3+a2ba2+b2=bb2+a2a2+b2=b

Hence, the correct answer is option B.

Page No 9.45:

Question 37:

If tanα=17,tanβ=13, then cos2α is equal to

(a) sin2β                                 (b) sin4β                                 (c) sin3β                                 (d) cos2β                                 
                                  
               

Answer:

It is given that tanα=17 and tanβ=13.

Now,

tan2β=2tanβ1-tan2β           =2×131-19           =2389           =34

tanα+2β=tanα+tan2β1-tanα tan2β                        =17+341-17×34                        =25282528                        =1

tanα+2β=1=tanπ4α+2β=π4α=π4-2β2α=π2-4βcos2α=cosπ2-4β=sin4β

cos2α=sin4β

Hence, the correct answer is option B.

Page No 9.45:

Question 38:

The value of cos248°-sin212° is

(a) 5+18                             (b) 5-18                             (c) 5+15                             (d) 5+122                            

Answer:

cos248°-sin212°=cos48°+12°cos48°-12°                      cosA+BcosA-B=cos2A-sin2B=cos60°cos36°=12×5+14=5+18                

Hence, the correct answer is option A.



View NCERT Solutions for all chapters of Class 12