Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions (Exercise 3.3) with step-by-step explanations and clear derivations based on the latest CBSE and NCERT syllabus. This exercise focuses on trigonometric ratios of acute angles, values of sine, cosine, and tangent functions, and the use of standard angles in radians and degrees. These solutions help students develop strong conceptual understanding and problem-solving skills useful for school exams, JEE, NDA, CUET, and other competitive exams. All answers are carefully prepared by experienced faculty for accuracy and easy learning. Click the print button to download study material and notes in PDF format.
Q.1 : Prove that $ \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} – \tan^2\frac{\pi}{4} = -\frac{1}{2} $
Solution
$$
\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} – \tan^2\frac{\pi}{4}
= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 – 1^2
$$
$$
= \frac{1}{4} + \frac{1}{4} – 1 = \frac{1}{2} – 1 = -\frac{1}{2}
$$
Hence proved.
Q.2 : Prove that $ 2\sin^2\frac{\pi}{6} + \csc^2\frac{7\pi}{6} \cos^2\frac{\pi}{3} = \frac{3}{2} $
Solution
$$
2\sin^2\frac{\pi}{6} + \csc^2\frac{7\pi}{6} \cos^2\frac{\pi}{3}
= 2\left(\frac{1}{2}\right)^2 + \csc^2\left(\pi + \frac{\pi}{6}\right)\left(\frac{1}{2}\right)^2
$$
Since $\csc(\pi + \theta) = \csc\theta$,
$$
= 2\cdot\frac{1}{4} + \csc^2\frac{\pi}{6}\cdot\frac{1}{4}
= \frac{1}{2} + \left(2\right)^2\cdot\frac{1}{4}
$$
$$
= \frac{1}{2} + 1 = \frac{3}{2}
$$
Hence proved.
Q.3 : Prove that $ \cot^2\frac{\pi}{6} + \csc\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6 $
Solution
We know: $\cot\frac{\pi}{6} = \sqrt{3}$, $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, and $\csc\frac{5\pi}{6} = \csc(\pi – \frac{\pi}{6}) = \csc\frac{\pi}{6} = 2$.
$$
\cot^2\frac{\pi}{6} + \csc\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6}
= (\sqrt{3})^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2
$$
$$
= 3 + 2 + 1 = 6
$$
Hence proved.
Q.4 : Prove that $ 2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10 $
Solution
We know: $\sin\frac{3\pi}{4} = \sin(\pi – \frac{\pi}{4}) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$,
$\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, and $\sec\frac{\pi}{3} = 2$.
$$
2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3}
= 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2(2)^2
$$
$$
= 1 + 1 + 8 = 10
$$
Hence proved.
Q.5.1 : Find the value of $\sin 75^\circ$
$$
\sin 75^\circ = \sin(45^\circ + 30^\circ)
$$
Using $\sin(x + y) = \sin x \cos y + \cos x \sin y$,
$$
\sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ
$$
$$
= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2}
= \frac{\sqrt{3} + 1}{2\sqrt{2}}
$$
Hence, $\sin 75^\circ = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$.
Q.5.2 : Find the value of $\tan 15^\circ$
$$
\tan 15^\circ = \tan(45^\circ – 30^\circ)
$$
Using $\tan(x – y) = \dfrac{\tan x – \tan y}{1 + \tan x \tan y}$,
$$
\tan 15^\circ = \frac{1 – \frac{1}{\sqrt{3}}}{1 + 1\cdot\frac{1}{\sqrt{3}}}
= \frac{\sqrt{3} – 1}{\sqrt{3} + 1}
$$
Rationalizing the denominator:
$$
\tan 15^\circ = \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{4 – 2\sqrt{3}}{2} = 2 – \sqrt{3}.
$$
Hence, $\tan 15^\circ = 2 – \sqrt{3}$
Q.6 : Prove that
$\cos\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-y\right)
-\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}-y\right)
=\sin(x+y)$
Solution
LHS =
$$
\cos\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-y\right)
-\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}-y\right)
$$
Multiply and divide by 2:
$$
=\frac{1}{2}\Big[2\cos\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-y\right)
-2\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}-y\right)\Big]
$$
Use identities
$2\cos A\cos B = \cos(A+B)+\cos(A-B)$
and
$-2\sin A\sin B = \cos(A+B)-\cos(A-B)$.
Then,
$
\text{LHS}=\frac{1}{2}\Big[\cos\big((\tfrac{\pi}{4}-x)+(\tfrac{\pi}{4}-y)\big)
+\cos\big((\tfrac{\pi}{4}-x)-(\tfrac{\pi}{4}-y)\big)
-\cos\big((\tfrac{\pi}{4}-x)+(\tfrac{\pi}{4}-y)\big)
+\cos\big((\tfrac{\pi}{4}-x)-(\tfrac{\pi}{4}-y)\big)\Big]
$
Simplify:
$$
=\frac{1}{2}\Big[2\cos\left(\frac{\pi}{2}-(x+y)\right)\Big]
=\cos\left(\frac{\pi}{2}-(x+y)\right)
$$
Since $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$,
$$
\text{LHS} = \sin(x+y) = \text{RHS}
$$
Hence proved.
Q.7 : Prove that$$
\frac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)}
=\left(\frac{1+\tan x}{1-\tan x}\right)^2$$
Solution
Use the identities:
$$
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}, \quad
\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
$$
Then,
$$
\frac{\tan(\tfrac{\pi}{4}+x)}{\tan(\tfrac{\pi}{4}-x)}
=\frac{\dfrac{1+\tan x}{1-\tan x}}{\dfrac{1-\tan x}{1+\tan x}}
$$
Simplify step by step:
$$
=\frac{1+\tan x}{1-\tan x}\times\frac{1+\tan x}{1-\tan x}
$$
$$
=\left(\frac{1+\tan x}{1-\tan x}\right)^2
$$
Hence proved.
Q.8 : Prove that $$
\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\frac{\pi}{2}+x\right)}=\cot^2 x$$
Solution
We know:
$\cos(\pi+x)=-\cos x$,
$\cos(-x)=\cos x$,
$\sin(\pi-x)=\sin x$,
$\cos\left(\frac{\pi}{2}+x\right)=-\sin x$.
$$
\text{LHS}=\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\frac{\pi}{2}+x\right)}$$
Now substitute:
$$
=\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}
$$
Simplify:
$$
=\frac{\cos^2 x}{\sin^2 x}
=\left(\frac{\cos x}{\sin x}\right)^2
=\cot^2 x
$$
Hence proved.
Q.9 : Prove that $$
\cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x)
\Big[\cot\left(\frac{3\pi}{2}-x\right)+\cot(2\pi+x)\Big]=1 $$
Solution
We know:
$\cos\left(\frac{3\pi}{2}+x\right)=\sin x$,
$\cos(2\pi+x)=\cos x$,
$\cot\left(\frac{3\pi}{2}-x\right)=\tan x$,
$\cot(2\pi+x)=\cot x$.
$$
\text{LHS}=\cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x)
\Big[\cot\left(\frac{3\pi}{2}-x\right)+\cot(2\pi+x)\Big] $$
Now substitute:
$$
=\sin x\cos x[\tan x+\cot x]
$$
Simplify stepwise:
$$
=\sin x\cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)
$$
$$
=\sin x\cos x\cdot\frac{\sin^2 x+\cos^2 x}{\sin x\cos x}
$$
$$
=1
$$
Hence proved.
Q.10 : Prove that $$
\sin((n+1)x)\sin((n+2)x)+\cos((n+1)x)\cos((n+2)x)=\cos x $$
Solution :
$$
\text{LHS}=\sin((n+1)x)\sin((n+2)x)+\cos((n+1)x)\cos((n+2)x) $$
Multiply and divide by 2:
$$
=\frac{1}{2}\Big[2\sin((n+1)x)\sin((n+2)x)
+2\cos((n+1)x)\cos((n+2)x)\Big]
$$
Use formulas:
$$
2\sin A\sin B=\cos(A-B)-\cos(A+B)
$$
and
$$
2\cos A\cos B=\cos(A-B)+\cos(A+B)
$$
Substitute:
$
\text{LHS}=\frac{1}{2}\Big[\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)\Big]
$
Simplify term by term:
$$
=\frac{1}{2}[2\cos((n+1)x-(n+2)x)]
$$
$$
=\frac{1}{2}[2\cos(-x)] = \cos x
$$
Hence proved.
Q.11 : Prove that $$
\cos\left(\frac{3\pi}{4} + x\right) – \cos\left(\frac{3\pi}{4} – x\right) = -\sqrt{2}\sin x $$
Solution
LHS
$$
= \cos\left(\frac{3\pi}{4} + x\right) – \cos\left(\frac{3\pi}{4} – x\right)
$$
Using the identity
$$
\cos A – \cos B = -2 \sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A – B}{2}\right),
$$
we get
$$
= -2 \sin\left(\frac{\frac{3\pi}{4} + x + \frac{3\pi}{4} – x}{2}\right)
\sin\left(\frac{\frac{3\pi}{4} + x – \frac{3\pi}{4} + x}{2}\right)
$$
$$
= -2 \sin\left(\frac{3\pi}{2} \times \frac{1}{2}\right) \sin\left(\frac{2x}{2}\right)
$$
$$
= -2 \sin\frac{3\pi}{4} \sin x
$$
Now,
$$
\sin\frac{3\pi}{4} = \sin\left(\pi – \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}
$$
Hence,
$$
\text{LHS} = -2 \times \frac{1}{\sqrt{2}} \sin x = -\sqrt{2}\sin x = \text{RHS}
$$
Hence proved.
Q.12 : Prove that $$
\sin^2 6x – \sin^2 4x = \sin 2x \sin 10x $$
Solution
LHS
$$
\sin^2 6x – \sin^2 4x = (\sin 6x + \sin 4x)(\sin 6x – \sin 4x)
$$
Using the identities
$\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A – B}{2}$
and
$\sin A – \sin B = 2\cos\frac{A + B}{2}\sin\frac{A – B}{2}$,
we get
$$
= [2\sin\frac{6x + 4x}{2}\cos\frac{6x – 4x}{2}]
[2\cos\frac{6x + 4x}{2}\sin\frac{6x – 4x}{2}]
$$
$$
= [2\sin 5x \cos x][2\cos 5x \sin x]
$$
$$
= 4\sin 5x \cos 5x \sin x \cos x
$$
$$
= (2\sin 5x \cos 5x) (2\sin x \cos x)
$$
Now, using $\sin 2A = 2\sin A \cos A$, we get
$$
\text{LHS} = \sin 10x \sin 2x = \text{RHS}
$$
Hence proved.
Q.13 : Prove that $$
\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x $$
Solution
LHS
$$
\cos^2 2x – \cos^2 6x = (\cos 2x + \cos 6x)(\cos 2x – \cos 6x)
$$
Using the identities
$\cos A + \cos B = 2\cos\frac{A + B}{2}\cos\frac{A – B}{2}$
and
$\cos A – \cos B = -2\sin\frac{A + B}{2}\sin\frac{A – B}{2}$,
we get
$$
= [2\cos 4x \cos(-2x)][-2\sin 4x \sin(-2x)]
$$
$$
= [2\cos 4x \cos 2x][2\sin 4x \sin 2x]
$$
$$
= (2\sin 4x \cos 4x)(2\sin 2x \cos 2x)
$$
Using $\sin 2A = 2\sin A \cos A$, we get
$$
\text{LHS} = \sin 8x \sin 4x = \text{RHS}
$$
Hence proved.
Q.14 : Prove that $$
\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x $$
Solution
LHS
$$
\sin 2x + 2\sin 4x + \sin 6x = (\sin 2x + \sin 6x) + 2\sin 4x
$$
Using $\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A – B}{2}$,
$$
= 2\sin\frac{8x}{2}\cos\frac{-4x}{2} + 2\sin 4x
$$
$$
= 2\sin 4x \cos 2x + 2\sin 4x
$$
$$
= 2\sin 4x(\cos 2x + 1)
$$
Now, $\cos 2x = 2\cos^2 x – 1$, so
$$
\text{LHS} = 2\sin 4x(2\cos^2 x – 1 + 1)
$$
$$
= 4\cos^2 x \sin 4x = \text{RHS}
$$
Hence proved.
Q.15 : Prove that $$
\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x – \sin 3x) $$
Solution :
LHS
$$
=\cot 4x (\sin 5x + \sin 3x)
$$
$$
= \frac{\cos 4x}{\sin 4x} (\sin 5x + \sin 3x)
$$
Using $\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A – B}{2}$,
$$
= \frac{\cos 4x}{\sin 4x}[2\sin 4x \cos x]
$$
$$
= 2\cos 4x \cos x
$$
Now, RHS
$$
=\cot x (\sin 5x – \sin 3x)
$$
$$
= \frac{\cos x}{\sin x} (\sin 5x – \sin 3x)
$$
Using $\sin A – \sin B = 2\cos\frac{A + B}{2}\sin\frac{A – B}{2}$,
$$
= \frac{\cos x}{\sin x}[2\cos 4x \sin x]
$$
$$
= 2\cos 4x \cos x
$$
Hence,
$$
\text{LHS} = \text{RHS}
$$
Hence proved.
Q.16 : Prove that $$
\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = -\frac{\sin 2x}{\cos 10x}$$
Solution
LHS
$$
= \frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x}
$$
Using the identities
$\cos A – \cos B = -2\sin\frac{A + B}{2}\sin\frac{A – B}{2}$
and
$\sin A – \sin B = 2\cos\frac{A + B}{2}\sin\frac{A – B}{2}$,
we get
$$
= \frac{-2\sin\frac{9x + 5x}{2}\sin\frac{9x – 5x}{2}}
{2\cos\frac{17x + 3x}{2}\sin\frac{17x – 3x}{2}}
$$
$$
= \frac{-2\sin 7x \sin 2x}{2\cos 10x \sin 7x}
$$
$$
= -\frac{\sin 2x}{\cos 10x} = \text{RHS}
$$
Hence proved.
Q.17 : Prove that $$
\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x $$
Solution
LHS
$$
= \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}
$$
Using
$\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A – B}{2}$
and
$\cos A + \cos B = 2\cos\frac{A + B}{2}\cos\frac{A – B}{2}$,
we get
$$
= \frac{2\sin 4x \cos x}{2\cos 4x \cos x}
$$
$$
= \frac{\sin 4x}{\cos 4x} = \tan 4x = \text{RHS}
$$
Hence proved.
18. Prove that
$$
\frac{\sin x – \sin y}{\cos x + \cos y} = \tan\frac{x – y}{2}
$$
Solution
LHS
$$
= \frac{\sin x – \sin y}{\cos x + \cos y}
$$
Using
$\sin A – \sin B = 2\cos\frac{A + B}{2}\sin\frac{A – B}{2}$
and
$\cos A + \cos B = 2\cos\frac{A + B}{2}\cos\frac{A – B}{2}$,
we get
$$
= \frac{2\cos\frac{x + y}{2}\sin\frac{x – y}{2}}
{2\cos\frac{x + y}{2}\cos\frac{x – y}{2}}
$$
$$
= \frac{\sin\frac{x – y}{2}}{\cos\frac{x – y}{2}}
= \tan\frac{x – y}{2} = \text{RHS}
$$
Hence proved.
19. Prove that
$$
\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x
$$
Solution
LHS
$$
= \frac{\sin x + \sin 3x}{\cos x + \cos 3x}
$$
Using
$\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A – B}{2}$
and
$\cos A + \cos B = 2\cos\frac{A + B}{2}\cos\frac{A – B}{2}$,
we get
$$
= \frac{2\sin 2x \cos x}{2\cos 2x \cos x}
$$
$$
= \frac{\sin 2x}{\cos 2x} = \tan 2x = \text{RHS}
$$
Hence proved.
20. Prove that
$$
\frac{\sin x – \sin 3x}{\sin^2 x – \cos^2 x} = 2\sin x
$$
Solution
LHS
$$
= \frac{\sin x – \sin 3x}{\sin^2 x – \cos^2 x}
$$
Using $\sin A – \sin B = 2\cos\frac{A + B}{2}\sin\frac{A – B}{2}$,
we get
$$
= \frac{2\cos 2x \sin(-x)}{\sin^2 x – \cos^2 x}
$$
$$
= \frac{-2\cos 2x \sin x}{-(\cos^2 x – \sin^2 x)}
$$
But $\cos^2 x – \sin^2 x = \cos 2x$, so
$$
\text{LHS} = \frac{2\cos 2x \sin x}{\cos 2x} = 2\sin x = \text{RHS}
$$
Hence proved.
