Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 3 Ex 3.3 — all 25 questions solved, covering direct substitution, sum and difference formulas, sum-to-product conversions, and multiple-angle identities.
This is the longest exercise in the chapter, but nearly every proof draws from the same small toolbox. Questions 1–5 are direct substitution using known values at π/6, π/4 and π/3. Questions 6–11 apply the sum/difference formulas for sin and cos (and the related \tan(\pi/4\pm x) results) directly. Questions 12–21 all convert a sum or difference of sines/cosines into a product — the four sum-to-product identities below are worth memorising cold, since roughly a third of this exercise depends on them. Questions 22–25 close with the double- and triple-angle formulas for tan, cos, and a triple-angle expansion in terms of \cos2x.
L.H.S. =\sin^2\dfrac{\pi}{6}+\cos^2\dfrac{\pi}{3}-\tan^2\dfrac{\pi}{4}
Substituting the standard values \sin\dfrac{\pi}{6}=\dfrac{1}{2}, \cos\dfrac{\pi}{3}=\dfrac{1}{2}, \tan\dfrac{\pi}{4}=1:
=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-1^2=\dfrac{1}{4}+\dfrac{1}{4}-1
=\dfrac{1}{2}-1=-\dfrac{1}{2} = R.H.S.
L.H.S. =2\sin^2\dfrac{\pi}{6}+\text{cosec}^2\dfrac{7\pi}{6}\cos^2\dfrac{\pi}{3}
\sin\dfrac{\pi}{6}=\dfrac{1}{2}\ \Rightarrow\ 2\sin^2\dfrac{\pi}{6}=2\times\dfrac{1}{4}=\dfrac{1}{2}
Since \dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}: \sin\dfrac{7\pi}{6}=-\sin\dfrac{\pi}{6}=-\dfrac{1}{2}, so \text{cosec}\dfrac{7\pi}{6}=-2 and \text{cosec}^2\dfrac{7\pi}{6}=4.
\cos\dfrac{\pi}{3}=\dfrac{1}{2}\ \Rightarrow\ \cos^2\dfrac{\pi}{3}=\dfrac{1}{4}
Therefore, L.H.S. =\dfrac{1}{2}+4\times\dfrac{1}{4}=\dfrac{1}{2}+1=\dfrac{3}{2} = R.H.S.
L.H.S. =\cot^2\dfrac{\pi}{6}+\text{cosec}\dfrac{5\pi}{6}+3\tan^2\dfrac{\pi}{6}
\cot\dfrac{\pi}{6}=\sqrt3\ \Rightarrow\ \cot^2\dfrac{\pi}{6}=3
Since \dfrac{5\pi}{6}=\pi-\dfrac{\pi}{6}: \sin\dfrac{5\pi}{6}=\sin\dfrac{\pi}{6}=\dfrac{1}{2}, so \text{cosec}\dfrac{5\pi}{6}=2.
\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}\ \Rightarrow\ \tan^2\dfrac{\pi}{6}=\dfrac{1}{3}\ \Rightarrow\ 3\tan^2\dfrac{\pi}{6}=1
Therefore, L.H.S. =3+2+1=6 = R.H.S.
L.H.S. =2\sin^2\dfrac{3\pi}{4}+2\cos^2\dfrac{\pi}{4}+2\sec^2\dfrac{\pi}{3}
Since \dfrac{3\pi}{4}=\pi-\dfrac{\pi}{4}: \sin\dfrac{3\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\ \Rightarrow\ 2\sin^2\dfrac{3\pi}{4}=2\times\dfrac{1}{2}=1
\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\ \Rightarrow\ 2\cos^2\dfrac{\pi}{4}=2\times\dfrac{1}{2}=1
\sec\dfrac{\pi}{3}=2\ \Rightarrow\ 2\sec^2\dfrac{\pi}{3}=2\times4=8
Therefore, L.H.S. =1+1+8=10 = R.H.S.
(i) Write 75^\circ=45^\circ+30^\circ.
\sin75^\circ=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ
=\dfrac{1}{\sqrt2}\times\dfrac{\sqrt3}{2}+\dfrac{1}{\sqrt2}\times\dfrac{1}{2}=\dfrac{\sqrt3+1}{2\sqrt2}
(ii) Write 15^\circ=45^\circ-30^\circ.
\tan15^\circ=\dfrac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}=\dfrac{1-\tfrac{1}{\sqrt3}}{1+\tfrac{1}{\sqrt3}}=\dfrac{\sqrt3-1}{\sqrt3+1}
Rationalising by multiplying numerator and denominator by (\sqrt3-1):
=\dfrac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}=\dfrac{4-2\sqrt3}{2}
L.H.S. is of the form \cos A\cos B-\sin A\sin B=\cos(A+B), with A=\dfrac{\pi}{4}-x and B=\dfrac{\pi}{4}-y.
L.H.S. =\cos\left[\left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}-y\right)\right]=\cos\left(\dfrac{\pi}{2}-x-y\right)
Using \cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta, with \theta=x+y:
=\sin(x+y) = R.H.S.
Using the tangent sum and difference formulas with \tan\dfrac{\pi}{4}=1:
\tan\left(\dfrac{\pi}{4}+x\right)=\dfrac{1+\tan x}{1-\tan x} and \tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}
Therefore, L.H.S. =\dfrac{1+\tan x}{1-\tan x}\div\dfrac{1-\tan x}{1+\tan x}=\dfrac{1+\tan x}{1-\tan x}\times\dfrac{1+\tan x}{1-\tan x}
=\left(\dfrac{1+\tan x}{1-\tan x}\right)^2 = R.H.S.
Simplify each factor using standard results: \cos(\pi+x)=-\cos x, \cos(-x)=\cos x, \sin(\pi-x)=\sin x, \cos\left(\dfrac{\pi}{2}+x\right)=-\sin x.
L.H.S. =\dfrac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}=\dfrac{-\cos^2x}{-\sin^2x}=\dfrac{\cos^2x}{\sin^2x}
=\cot^2x = R.H.S.
Simplify each factor:
\cos\left(\dfrac{3\pi}{2}+x\right)=\sin x, \cos(2\pi+x)=\cos x
\cot\left(\dfrac{3\pi}{2}-x\right)=\tan x (since \cos\left(\tfrac{3\pi}{2}-x\right)=-\sin x and \sin\left(\tfrac{3\pi}{2}-x\right)=-\cos x, giving \cot=\tfrac{-\sin x}{-\cos x}=\tan x)
\cot(2\pi+x)=\cot x (period \pi, and 2\pi is a multiple of \pi)
So the bracket becomes \tan x+\cot x=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=\dfrac{\sin^2x+\cos^2x}{\sin x\cos x}=\dfrac{1}{\sin x\cos x}
Therefore, L.H.S. =(\sin x)(\cos x)\times\dfrac{1}{\sin x\cos x}=1 = R.H.S.
L.H.S. is of the form \cos A\cos B+\sin A\sin B=\cos(A-B), with A=(n+1)x and B=(n+2)x.
L.H.S. =\cos\left[(n+1)x-(n+2)x\right]=\cos(-x)
Since \cos(-x)=\cos x:
L.H.S. =\cos x = R.H.S.
Using \cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}, with C=\dfrac{3\pi}{4}+x and D=\dfrac{3\pi}{4}-x:
\dfrac{C+D}{2}=\dfrac{3\pi}{4} and \dfrac{C-D}{2}=x
L.H.S. =-2\sin\dfrac{3\pi}{4}\sin x
Since \sin\dfrac{3\pi}{4}=\sin\left(\pi-\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}:
L.H.S. =-2\times\dfrac{1}{\sqrt2}\times\sin x=-\sqrt2\sin x = R.H.S.
We use the identity \sin^2A-\sin^2B=\sin(A+B)\sin(A-B), with A=6x and B=4x.
L.H.S. =\sin(6x+4x)\sin(6x-4x)=\sin10x\sin2x
=\sin2x\sin10x = R.H.S.
Using \cos^2A-\cos^2B=\sin^2B-\sin^2A=\sin(A+B)\sin(B-A), with A=2x and B=6x:
L.H.S. =\sin(2x+6x)\sin(6x-2x)=\sin8x\sin4x
=\sin4x\sin8x = R.H.S.
Group the first and third terms and apply the sum-to-product formula:
\sin2x+\sin6x=2\sin\dfrac{2x+6x}{2}\cos\dfrac{6x-2x}{2}=2\sin4x\cos2x
So L.H.S. =2\sin4x\cos2x+2\sin4x=2\sin4x(\cos2x+1)
Using \cos2x=2\cos^2x-1\ \Rightarrow\ \cos2x+1=2\cos^2x:
L.H.S. =2\sin4x\times2\cos^2x=4\cos^2x\sin4x = R.H.S.
Using sum-to-product: \sin5x+\sin3x=2\sin4x\cos x and \sin5x-\sin3x=2\cos4x\sin x.
L.H.S. =\cot4x\times2\sin4x\cos x=\dfrac{\cos4x}{\sin4x}\times2\sin4x\cos x=2\cos4x\cos x
R.H.S. =\cot x\times2\cos4x\sin x=\dfrac{\cos x}{\sin x}\times2\cos4x\sin x=2\cos4x\cos x
Since L.H.S. = R.H.S. =2\cos4x\cos x,
Numerator: \cos9x-\cos5x=-2\sin\dfrac{9x+5x}{2}\sin\dfrac{9x-5x}{2}=-2\sin7x\sin2x
Denominator: \sin17x-\sin3x=2\cos\dfrac{17x+3x}{2}\sin\dfrac{17x-3x}{2}=2\cos10x\sin7x
L.H.S. =\dfrac{-2\sin7x\sin2x}{2\cos10x\sin7x}=-\dfrac{\sin2x}{\cos10x} = R.H.S.
Numerator: \sin5x+\sin3x=2\sin4x\cos x
Denominator: \cos5x+\cos3x=2\cos4x\cos x
L.H.S. =\dfrac{2\sin4x\cos x}{2\cos4x\cos x}=\dfrac{\sin4x}{\cos4x}=\tan4x = R.H.S.
Numerator: \sin x-\sin y=2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}
Denominator: \cos x+\cos y=2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}
L.H.S. =\dfrac{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}=\dfrac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}}=\tan\dfrac{x-y}{2} = R.H.S.
Numerator: \sin x+\sin3x=2\sin2x\cos x
Denominator: \cos x+\cos3x=2\cos2x\cos x
L.H.S. =\dfrac{2\sin2x\cos x}{2\cos2x\cos x}=\dfrac{\sin2x}{\cos2x}=\tan2x = R.H.S.
Numerator: \sin x-\sin3x=2\cos\dfrac{x+3x}{2}\sin\dfrac{x-3x}{2}=2\cos2x\sin(-x)=-2\cos2x\sin x
Denominator: \sin^2x-\cos^2x=-(\cos^2x-\sin^2x)=-\cos2x
L.H.S. =\dfrac{-2\cos2x\sin x}{-\cos2x}=2\sin x = R.H.S.
Group the outer terms of the numerator: \cos4x+\cos2x=2\cos3x\cos x.
So numerator =2\cos3x\cos x+\cos3x=\cos3x(2\cos x+1).
Similarly, group the outer terms of the denominator: \sin4x+\sin2x=2\sin3x\cos x.
So denominator =2\sin3x\cos x+\sin3x=\sin3x(2\cos x+1).
L.H.S. =\dfrac{\cos3x(2\cos x+1)}{\sin3x(2\cos x+1)}=\dfrac{\cos3x}{\sin3x}=\cot3x = R.H.S.
Since 3x=x+2x, applying the cotangent addition formula:
\cot3x=\cot(x+2x)=\dfrac{\cot x\cot2x-1}{\cot2x+\cot x}
Multiplying both sides by (\cot x+\cot2x):
\cot3x(\cot x+\cot2x)=\cot x\cot2x-1
\cot3x\cot x+\cot3x\cot2x=\cot x\cot2x-1
Rearranging:
\cot x\cot2x-\cot2x\cot3x-\cot3x\cot x=1 = R.H.S.
Write 4x=2(2x), so \tan4x=\dfrac{2\tan2x}{1-\tan^22x}, where \tan2x=\dfrac{2\tan x}{1-\tan^2x}.
Squaring: \tan^22x=\dfrac{4\tan^2x}{(1-\tan^2x)^2}
So 1-\tan^22x=\dfrac{(1-\tan^2x)^2-4\tan^2x}{(1-\tan^2x)^2}
Expanding the numerator: (1-\tan^2x)^2-4\tan^2x=1-2\tan^2x+\tan^4x-4\tan^2x=1-6\tan^2x+\tan^4x
So 1-\tan^22x=\dfrac{1-6\tan^2x+\tan^4x}{(1-\tan^2x)^2}
Therefore:
\tan4x=\dfrac{2\times\dfrac{2\tan x}{1-\tan^2x}}{\dfrac{1-6\tan^2x+\tan^4x}{(1-\tan^2x)^2}}=\dfrac{4\tan x}{1-\tan^2x}\times\dfrac{(1-\tan^2x)^2}{1-6\tan^2x+\tan^4x}
=\dfrac{4\tan x(1-\tan^2x)}{1-6\tan^2x+\tan^4x} = R.H.S.
Write 4x=2(2x), so using \cos2\theta=1-2\sin^2\theta with \theta=2x:
\cos4x=1-2\sin^22x
Since \sin2x=2\sin x\cos x, we have \sin^22x=4\sin^2x\cos^2x.
Therefore:
\cos4x=1-2\times4\sin^2x\cos^2x=1-8\sin^2x\cos^2x = R.H.S.
Write 6x=3(2x), so using the triple-angle formula \cos3\theta=4\cos^3\theta-3\cos\theta with \theta=2x:
\cos6x=4\cos^32x-3\cos2x
Let c=\cos^2x, so \cos2x=2c-1 (using \cos2x=2\cos^2x-1).
Then \cos^32x=(2c-1)^3=8c^3-12c^2+6c-1, so 4\cos^32x=32c^3-48c^2+24c-4.
Also 3\cos2x=3(2c-1)=6c-3.
Therefore:
\cos6x=(32c^3-48c^2+24c-4)-(6c-3)=32c^3-48c^2+18c-1
Substituting back c=\cos^2x:
\cos6x=32\cos^6x-48\cos^4x+18\cos^2x-1 = R.H.S.
Expert CBSE Coaching · Class 9–12