Class 11 Maths Ncert Solutions Ex 3.3

Class 11 Maths Ncert Solutions Chapter 3 Ex 3.3

Q.1: Prove:

sin2π6+cos2π3tan2π4=12

 

Sol:

Now, taking L.H.S.

sin2π6+cos2π3tan2π4:

= (12)2+(12)2(1)2

= 14+141 = 12

= R.H.S.

 

 

Q.2: Prove:

2sin2π6+cosec27π6cos2π3=32

 

Sol:

Now, taking L.H.S.

2sin2π6+cosec27π6cos2π3 :

= 2(12)2+cosec2(π+π6)(12)2

= 2×14+(cosecπ6)2(14)

= 12+(2)2(14)

= 12+44

= 12+1

= 32

= R.H.S.

 

 

Q.3: Prove:

cot2π6+cosec5π6+3tan2π6=6

 

Sol:

Taking L.H.S.

cot2π6+cosec5π6+3tan2π6 :

= (3)2+cosec(ππ6)+3(13)2

= 3+cosecπ6+3×13

= 3 + 2 + 1 = 6

= R.H.S.

 

 

Q.4: Prove:

2sin23π4+2cos2π4+2sec2π3=10

 

Sol:

Now, taking L.H.S.

2sin23π4+2cos2π4+2sec2π3 :

= 2{sin(ππ4)}2+2(12)2+2(2)2

= 2{sinπ4}2+2×12+8

= 2(12)2 + 1 + 8

= 1 + 1 + 8 = 10

= R.H.S.

 

 

Q.5: Calculate the value of:

(i).  sin75

 

(ii).  tan15

 

Sol:

(i).  sin75:

= sin(45+30)

= sin45cos30+cos45sin30

Since, [sin (x + y) = sin x cos y + cos x sin y]

= (12)(32)+(12)(12)

= 322+122

= 3+122

 

(ii).  tan15:

= tan(4530)

= tan45tan301+tan45tan30

Since, [tan (x – y) = tanxtany1+tanxtany]

= 1131+1(13)

= 3133+13

= 313+1

= (31)2(3+1)(31)

= 3+123(3)2(1)2

= 42331

= 23

 

 

Q.6:Prove:

cos(π4x)cos(π4y)sin(π4x)sin(π4y)=sin(x+y)

 

Sol:

Now, taking L.H.S.

cos(π4x)cos(π4y)sin(π4x)sin(π4y):

= 12[2cos(π4x)cos(π4y)]+12[2sin(π4x)sin(π4y)]

=12[cos{(π4x)+(π4y)}+cos{(π4x)(π4y)}]+12[cos{(π4x)+(π4y)}cos{(π4x)(π4y)}]

Since, [2cos A cos B = cos (A + B) + cos (A – B)]

And, [2sin A sin B = cos (A + B) – cos (A – B)]

= 2×12[cos{(π4x)+(π4y)}]

= cos[π2(x+y)]

= sin (x + y)

= R.H.S.

 

 

Q.7: Prove:

tan(π4+x)tan(π4x)=(1+tanx1tanx)2

 nbsp;

Sol:

Since, tan (A + B)= tanA+tanB1tanAtanB

And, tan (A – B) = tanAtanB1+tanAtanB

Now, taking L.H.S.

tan(π4+x)tan(π4x):

= (tanπ4+tanx1tanπ4tanx)(tanπ4tanx1+tanπ4tanx)

= (1+tanx1tanx)(1tanx1+tanx)

= (1+tanx1tanx)2

= R.H.S.

 

 

Q.8: Prove:

cos(π+x)cos(x)sin(πx)cos(π2+x)=cot2x

 

Sol:

Now, taking L.H.S.

cos(π+x)cos(x)sin(πx)cos(π2+x):

= [cosx][cosx](sinx)(sinx)

= cos2xsin2x = cot2x

= R.H.S.

 

 

Q.9: Prove:

cos(3π2+x)cos(2π+x)[cot(3π2x)+cot(2π+x)]=1

 

Sol:

Now, taking L.H.S.

cos(3π2+x)cos(2π+x)[cot(3π2x)+cot(2π+x)]:

= sinxcosx[tanx+cotx]

= sinxcosx(sinxcosx+cosxsinx)

= (sinxcosx)[sin2x+cos2xsinxcosx]

= sin2x+cos2x = 1

= R.H.S.

 

 

Q.10: Prove:

sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx

 

Sol:

Now, taking L.H.S.

sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x:

=12[2sin(n+1)xsin(n+2)x+2cos(n+1)xcos(n+2)x]

=12[cos{(n+1)x(n+2)x}cos{(n+1)x+(n+2)x}+cos{(n+1)x+(n+2)x}+cos{(n+1)x(n+2)x}]

Since, [2sin A sin B = cos (A + B) – cos (A – B)]

And, [2cos A cos B = cos (A + B) + cos (A – B)]

= 12×2cos{(n+1)x(n+2)x}

= cos(x) = cos x

= R.H.S.

 

 

Q.11 Prove:

cos(3π4+x)cos(3π4x)=2sinx

 

Sol:

Since, cos A – cos B = 2sin(A+B2)sin(AB2)

Now, taking L.H.S.

cos(3π4+x)cos(3π4x):

= 2sin{(3π4+x)+(3π4x)2}sin{(3π4+x)(3π4x)2}

= 2sin(3π4)sinx

= 2sin(ππ4)sinx

= 2sinπ4sinx

= 2×12×sinx

= 2sinx

= R.H.S.

 

 

Q.12: Prove:

sin26xsin24x=sin2xsin10x

 

Sol:

Since, sin A + sin B = 2sin(A+B2)cos(AB2)

And, sin A – sin B = 2cos(A+B2)sin(AB2)

Now, taking L.H.S.

sin26xsin24x:

= (sin6x+sin4x)(sin6xsin4x)

= [2sin(6x+4x2)cos(6x4x2)][2cos(6x+4x2)sin(6x4x2)]

= (2sin 5x cos x) (2cos 5x sin x)

= (2 sin 5x cos 5x) (2 cos x sin x)

= sin 10x sin 2x

= R.H.S.

 

 

Q.13: Prove:

cos22xcos26x=sin4xsin8x

 

Sol:

Since, cos A + cos B = 2cos(A+B2)cos(AB2)

And, cos A – cos B = 2sin(A+B2)sin(AB2)

Now, taking L.H.S.

cos22xcos26x:

= (cos 2x + cos 6x)  (cos 2x – cos 6x)

= [2cos(2x+6x2)cos(2x6x2)][2sin(2x+6x2)sin(2x6x2)]

= [2cos 4x cos (-2x)]   [ -2sin 4x sin ( -2x)]

= [2cos 4x cos 2x]  [-2sin 4x ( -sin2x)]

= [2sin 4x cos 4x]  [2sin 2x cos 2x]

= sin 8x sin 4x

= R.H.S.

 

 

Q.14:Prove:

sin2x+2sin4x+sin6x=4cos2xsin4x

 

Sol:

Now, taking L.H.S.

sin 2x + 2sin 4x + sin 6x:

= [sin 2x + sin 6x] + 2 sin 4x

= [2sin(2x+6x2)(2x6x2)]+2sin4x

Since, sin A + sin B = 2sin(A+B2)cos(AB2)

= 2sin 4x cos(-2x) + 2sin 4x

= 2sin 4x cos 2x + 2sin 4x

= 2sin 4x (cos 2x + 1)

= 2sin4x(2cos2x1+1)

= 2sin4x(2cos2x)

= 4cos2xsin4x

= R.H.S.

 

 

Q.15: Prove:

cot4x(sin5x+sin3x)=cotx(sin5xsin3x)

 

Sol:

Now, taking L.H.S.

cot 4x (sin 5x + sin 3x):

= cos4xsin4x[2sin(5x+3x2)cos(5x3x2)]

Since, sin A – sin B = 2cos(A+B2)sin(AB2)

= (cos4xsin4x)[2sin4xcosx]

=2cos 4x cos x . . . . . . . . . . . . . . . (1)

Now, taking R.H.S.

cot x (sin 5x – sin 3x):

= cosxsinx[2cos(5x+3x2)sin(5x3x2)]

sinAsinB=2cos(A+B2)sin(AB2)

= cosxsinx[2cos4xsinx]

= 2 cos 4x cos x . . . . . . . . . . . . . . . . . . . . (2)

From equation (1) and (2):

L.H.S. = R.H.S.

 

 

Q.16: Prove:

cos9xcos5xsin17xsin3x=sin2xcos10x

 

Sol:

Since, cos A – cos B = 2sin(A+B2)sin(AB2)

And, sin A – sin B = 2cos(A+B2)sin(AB2)

Now, taking L.H.S.

cos9xcos5xsin17xsin3x:

= 2sin(9x+5x2)sin(9x5x2)2cos(17x+3x2)sin(17x3x2)

= 2sin7xsin2x2cos10xsin7x

= sin2xcos10x

= R.H.S.

 

 

Q.17: Prove:

sin5x+sin3xcos5x+cos3x=tan4x

 

Sol:

Since, sin A + sin B = 2sin(A+B2)cos(AB2)

And, cos A + cos B = 2cos(A+B2)cos(AB2)

Now, taking L.H.S.

sin5x+sin3xcos5x+cos3x:

= 2sin(5x+3x2)cos(5x3x2)2cos(5x+3x2)cos(5x3x2)

= 2sin4xcosx2cos4xcosx = tan 4x

=R.H.S.

 

 

Q.18: Prove:

sinxsinycosx+cosy=tanxy2

 

Sol:

Since, cos A + cos B = 2cos(A+B2)cos(AB2)

And, sin A – sin B = 2cos(A+B2)sin(AB2)

Now taking L.H.S.

sinxsinycosx+cosy:

= 2cos(x+y2)sin(xy2)2cos(x+y2)cos(xy2)

= sin(xy2)cos(xy2)

= tan(xy2)

= R.H.S.

 

 

Q.19: Prove:

sinx+sin3xcosx+cos3x=tan2x

 

Sol::

Since, sin A + sin B = 2sin(A+B2)cos(AB2)

And, cos A + cos B = 2cos(A+B2)cos(AB2)

Now, taking L.H.S.

sinx+sin3xcosx+cos3x:

= 2sin(x+3x2)cos(x3x2)2cos(x+3x2)cos(x3x2)

= sin2xcos2x = tan 2x

= R.H.S.

 

 

Q.20: Prove:

sinxsin3xsin2xcos2x=2sinx

 

Answer:

Since, sin A – sin B = 2cos(A+B2)sin(AB2)

cos2Asin2A=cos2A

Now, taking L.H.S.

sinxsin3xsin2xcos2x:

= 2cos(x+3x2)sin(x3x2)cos2x

= 2cos2xsin(x)cos2x

= 2x(sinx) = 2 sin x

= R.H.S.

 

 

Q.21: Prove:

cos4x+cos3x+cos2xsin4x+sin3x+sin2x=cot3x

 

Sol:

Taking L.H.S.

cos4x+cos3x+cos2xsin4x+sin3x+sin2x:

= (cos4x+cos2x)+cos3x(sin4x+sin2x)+sin3x

= 2cos(4x+2x2)cos(4x2x2)+cos3x2sin(4x+2x2)cos(4x2x2)+sin3x

sinA+sinB=2sin(A+B2)cos(AB2)cosA+cosB=2cos(A+B2)cos(AB2)

= 2cos3xcosx+cos3x2sin3xcosx+sin3x

= cos3x(2cosx+1)sin3x(2cosx+1) = cot 3x

= R.H.S.

 

 

Q.22: Prove:

cotxcot2xcot2xcot3xcot3xcotx=1

 

Sol:

Now, taking L.H.S.

cot x cot 2x – cot 2x cot 3x – cot 3x cot x :

= cotxcot2xcot3x(cot2x+cotx)

= cotxcot2xcot(2x+x)(cot2x+cotx)

= cotxcot2x[cot2xcotx1cotx+cot2x](cot2x+cotx)

= [cot(A+B)=cotAcotB1cotA+cotB]

= cotxcot2x(cot2xcotx1) = 1

= R.H.S

 

 

Q.23: Prove:

tan4x=4tanx(1tan2x)16tan2x+tan4x

 

Sol:

Since, tan 2A = 2tanA1tan2A

Now, taking L.H.S.

tan 4x :

= tan2(2x)

= 2tan2x1tan2(2x)

= 2(2tanx1tan2x)1(2tanx1tan2x)2

= (4tanx1tan2x)14tan2x(1tan2x)2

=  (4tanx1tan2x)(1tan2x)24tan2x(1tan2x)2

= 4tanx(1tan2x)(1tan2x)24tan2x

= 4tanx(1tan2x)1+tan4x2tan2x4tan2x