Ex 3.3 Trigonomety ncert maths solution class 11

  EXERCISE 3.3(Trigonometric Function)

Ex 3.3 Trigonomety ncert maths solution class 11

Prove that:(Ex 3.3 Trigonomety ncert maths solution class 11)

Question 1: \sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}=-\frac{1}{2}

Solution: L.H.S. =\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}

=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-(1)^2

=1 / 4+1 / 4-1

=\frac{1+1-4}{4}=-\frac{2}{4}

=-1 / 2 =\text { RHS }

Question 2: 2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2}

Solution: \text { L.H.S. }=2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}

=2\left(\frac{1}{2}\right)^2+\operatorname{cosec}^2\left(\pi+\frac{\pi}{6}\right)\left(\frac{1}{2}\right)^2

=2 \times \frac{1}{4}+\left(\operatorname{cosec^2} \frac{\pi}{6}\right)\left(\frac{1}{4}\right)

=\frac{1}{2}+(-2)^2\left(\frac{1}{4}\right)

=1 / 2+4 / 4

=1 / 2+1

=3 / 2=\text { RHS }

Question 3: \cot ^2 \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}=6

Solution: L.H.S. =\cot ^2 \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}

=(\sqrt{3})^2+\operatorname{cosec}\left(\pi-\frac{\pi}{6}\right)+3\left(\frac{1}{\sqrt{3}}\right)^2

=3+\operatorname{cosec} \frac{\pi}{6}+3 \times \frac{1}{3}

=3+2+1

=6 =\text { RHS }

Question 4: 2 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3}=10

Solution: L.H.S =2 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3}

=2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^2+2\left(\frac{1}{\sqrt{2}}\right)^2+2(2)^2

=2\left\{\sin \frac{\pi}{4}\right\}^2+2 \times \frac{1}{2}+8

=2\left(\frac{1}{\sqrt{2}}\right)^2+1+8

=1+1+8

=10 =\text { RHS }

Question 5: Find the value of:

(i) \sin 75^{\circ}

(ii) \tan 15^{\circ}

Solution: (i) \sin 75^{\circ}

=\sin \left(45^{\circ}+30^{\circ}\right)

Using the formula

[\sin (x+y)=\sin x \cos y+\cos x \sin y]

=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}

=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)

=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}

=\frac{\sqrt{3}+1}{2 \sqrt{2}}

(ii) \tan 15^{\circ}

=\tan \left(45^{\circ}-30^{\circ}\right)

Using formula

\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}

=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}

Substituting the values

=\frac{1-\frac{1}{\sqrt{3}}}{1+1\left(\frac{1}{\sqrt{3}}\right)}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}

=\frac{\sqrt{3}-1}{\sqrt{3}+1}

Multiply by \sqrt{3}-1 in N^r and D^r

=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}

=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}

=\frac{3+1-2 \sqrt{3}}{(\sqrt{3})^2-(1)^2}

=\frac{4-2 \sqrt{3}}{3-1}=\frac{2(2-\sqrt{3})}{2}

=2-\sqrt{3}

Prove the following:

Question 6: \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)

Solution: LHS =\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)

We know the formula

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

= \cos(\frac{\pi}{4}-x+\frac{\pi}{4}-y)

=\cos(\frac{\pi}{2}-(x+y))

=\sin(x+y)=RHS

Question 7: \frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^2

Solution: \text { .L.H.S. }=\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}

By using the formula

\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \text { and } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}

So we get

=\frac{\left(\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}\right)}{\left(\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right)}

= \frac{\left(\frac{1+\tan x}{1-\tan x}\right)}{\left(\frac{1-\tan x}{1+\tan x}\right)}

= \left(\frac{1+\tan x}{1-\tan x}\right)^2= \text { RHS }

Question 8: \frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^2 x

Solution: \text { L.H.S. }=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}

=\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}

=\frac{-\cos ^2 x}{-\sin ^2 x}

=\cot ^2 \mathrm{x} ={RHS}

Question 9: \cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1

Solution: \text { L.H.S. }=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]

=\sin \mathrm{x} \cos \mathrm{x}(\tan \mathrm{x}+\cot \mathrm{x})

=\sin x.\cos x(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})

=\sin x\cos x(\frac{\sin^2x+\cos^2 x}{\sin x\cos x})

=1=RHS

Question 10: \sin(n+1)x\sin(n+2)x+\cos(n+1)x\cos(n+2)x=\cos x

Solution:LHS = \sin(n+1)x\sin(n+2)x+\cos(n+1)x\cos(n+2)x

=\cos(n+1)x\cos(n+2)x+\sin(n+1)x\sin(n+2)x

Using formula

\cos (x-y)=\cos x\cos y+\sin x\sin y

Hence,

=\cos[(n+1)x-(n+2)x]

=\cos[nx+x-nx-2x]

=\cos(-x)

=\cos x=RHS

Question 11: \cos(\frac{3\pi}{4}+x)-\cos(\frac{3\pi}{4}-x)=-\sqrt{2}\sin x

Solution: LHS=\cos(\frac{3\pi}{4}+x)-\cos(\frac{3\pi}{4}-x)

Using formula

\cos (x-y)=\cos x\cos y+\sin x\sin y

\cos (x+y)=\cos x\cos y-\sin x\sin y

NOW,

=(\cos \frac{3\pi}{4}\cos x-\sin \frac{3\pi}{4}\sin x)-(\cos \frac{3\pi}{4}\cos x+\sin \frac{3\pi}{4}\sin x)

=\cos \frac{3\pi}{4}\cos x-\sin \frac{3\pi}{4}\sin x-\cos \frac{3\pi}{4}\cos x-\sin \frac{3\pi}{4}\sin x

=-2\sin \frac{3\pi}{4}\sin x

=-2(\frac{1}{\sqrt{2}})\sin x

=-\sqrt{2}\sin x=RHS

Question 12: \sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x

Solution: \text { L.H.S. }=\sin ^2 6 x-\sin ^2 4 x

Using the formula

\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

So we get

=(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x)

=\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right]\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \cdot \sin \left(\frac{6 x-4 x}{2}\right)\right]

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

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

=\sin 10 \mathrm{x} \sin 2 \mathrm{x}=\text { RHS }

Question 13: \cos ^2 2 x-\cos ^2 6 x=\sin 4 x \sin 8 x

Solution: \text { L.H.S. }=\cos ^2 2 x-\cos ^2 6 x

Using the formula

\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

So we get

=(\cos 2 x+\cos 6 x)(\cos 2 x-6 x)

=\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \frac{(2 x-6 x)}{2}\right]

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

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

=(2 \sin 4 x \cos 4 x)(2 \sin 2 x \cos 2 x)

=\sin 8 x \sin 4 x=\text { RHS }

Question 14: \sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^2 x \sin 4 x

Solution: \text { L.H.S. }=\sin 2 x+2 \sin 4 x+\sin 6 x

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

Using the formula

\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

=\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]+2 \sin 4 x

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

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

Taking common terms

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

Using the formula

=2 \sin 4 x\left(2 \cos ^2 x-1+1\right)

=2 \sin 4 x\left(2 \cos ^2 x\right)

=4 \cos ^2 x \sin 4

=\text { R.H.S. }

Question 15: \cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)

Solution: \text { LHS }=\cot 4 x(\sin 5 x+\sin 3 x)

=\frac{\cos 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]

Using the formula

sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) .

=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]

=2 \cos 4 x \cos \mathrm{x}

Similarly

\text { R.H.S. }=\cot x(\sin 5 x-\sin 3 x)

=\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right]

Using the formula

\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]

=2 \cos 4 \mathrm{x} \cos \mathrm{x}

Hence, LHS = RHS.

Question 16: \frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}

Solution: \text { L.H.S }=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}

Using the formula

\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

=\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \cdot \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 x+3 x}{2}\right) \cdot \sin \left(\frac{17 x-3 x}{2}\right)}

=\frac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin 7 x}

=-\frac{\sin 2 x}{\cos 10 x}

=\text { RHS }

Question 17: \frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x

Solution: \text { L.H.S. }=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}

Using the formula

\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

=\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}

=\frac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x}

=\frac{\sin 4 x}{\cos 4 x}

=\tan 4 x =\text { RHS }

Question 18: \frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}

Solution: \text { L.H.S. }=\frac{\sin x-\sin y}{\cos x+\cos y}

Using the formula

\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

=\frac{2 \cos \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)}

=\frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)}

=\tan \left(\frac{x-y}{2}\right)=\text { RHS }

Question 19: \frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x

Solution: L.H.S. =\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}

Using the formula

\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}

=\frac{\sin 2 x}{\cos 2 x}

=\tan 2 \mathrm{x}= RHS

Question 20: \frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}=2 \sin x

Solution: \text { L.H.S. }=\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}

Using the formula

\sin A-\sin \mathrm{B}=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)

\cos ^2 A-\sin ^2 A=\cos 2 A

=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}

=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}

=-2(-\sin x)

=2 \sin x = RHS

Question 21: \frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x

Solution: \text { L.H.S. }=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}

=\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x}

Using the formula

\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

=\frac{2 \cos \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\cos 3 x}{2 \sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\sin 3 x}

=\frac{2 \cos 3 x \cos x+\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x}

=\frac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)}

=\cot 3 x = RHS

Question 22: \cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1

Solution: \text { LHS }=\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x

=\cot x \cot 2 x-\cot 3 x(\cot 2 x+\cot x)

=\cot x \cot 2 x-\cot (2 x+x)(\cot 2 x+\cot x)

Using the formula

\cot (\mathrm{A}+\mathrm{B})=\frac{\cot \mathrm{A} \cot \mathrm{B}-1}{\cot \mathrm{A}+\cot \mathrm{B}}

=\cot x \cot 2 \mathrm{x}-\left[\frac{\cot 2 \mathrm{x} \cot \mathrm{x}-1}{\cot x+\cot 2 \mathrm{x}}\right](\cot 2 \mathrm{x}+\cot \mathrm{x})

=\cot x \cot 2 x-(\cot 2 x \cot x-1)

=1 = RHS

Question 23: \tan 4 x=\frac{4 \tan x\left(1-\tan ^2 x\right)}{1-6 \tan ^2 x+\tan ^4 x}

Solution: \text { LHS }=\tan 4 x=\tan 2(2 x)

By using the formula

\tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}

=\frac{2 \tan 2 x}{1-\tan ^2(2 x)}

=\frac{2\left(\frac{2 \tan x}{1-\tan ^2 x}\right)}{1-\left(\frac{2 \tan x}{1-\tan ^2 x}\right)^2}

=\frac{\left(\frac{4 \tan x}{1-\tan ^2 x}\right)}{\left[1-\frac{4 \tan ^2 x}{\left(1-\tan ^2 x\right)^2}\right]}

Taking LCM

=\frac{\left(\frac{4 \tan x}{1-\tan ^2 x}\right)}{\left[\frac{\left(1-\tan ^2 x\right)^2-4 \tan ^2 x}{\left(1-\tan ^2 x\right)^2}\right]}

=\frac{4 \tan x\left(1-\tan ^2 x\right)}{\left(1-\tan ^2 x\right)^2-4 \tan ^2 x}

=\frac{4 \tan x\left(1-\tan ^2 x\right)}{1+\tan ^4 x-2 \tan ^2 x-4 \tan ^2 x}

=\frac{4 \tan x\left(1-\tan ^2 x\right)}{1-6 \tan ^2 x+\tan ^4 x} =RHS

Question 24: \cos 4 x=1-8 \sin ^2 x \cos ^2 x

Solution: \text { LHS }=\cos 4 \mathrm{x}

=\cos 2(2 x)

Using the formula

\cos 2 A=1-2 \sin ^2 A

=1-2 \sin ^2 2 x

Again by using the formula

\sin 2 A=2 \sin A \cos A

=1-2(2 \sin x \cos x)^2

=1-8 \sin ^2 x \cos ^2 x = R.H.S.

Question 25: \cos 6 x=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1

Solution: L.H.S. =\cos 6 x

=\cos 3(2 x)

Using the formula

\cos 3 A=4 \cos ^3 A-3 \cos A

=4 \cos ^3 2 x-3 \cos 2 x

Again by using formula

\cos 2 x=2 \cos ^2 x-1

=4\left[\left(2 \cos ^2 x-1\right)^3-3\left(2 \cos ^2 x-1\right)\right.

=4\left[\left(2 \cos ^2 x\right)^3-(1)^3-3\left(2 \cos ^2 x\right)^2+3\left(2 \cos ^2 x\right)\right]-6 \cos ^2 x+3

=4\left[8 \cos ^6 x-1-12 \cos ^4 x+6 \cos ^2 x\right]-6 \cos ^2 x+3

By multiplication

=32 \cos ^6 x-4-48 \cos ^4 x+24 \cos ^2 x-6 \cos ^2 x+3

=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1=R.H.S.


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