May 2026

If sinθ + cosθ = 1, Find sin2θ If \( \sin\theta+\cos\theta=1 \), Find the Value of \( \sin2\theta \) Question If \[ \sin\theta+\cos\theta=1, \] then the value of \[ \sin2\theta \] is (a) \(1\) (b) \(\frac12\) (c) \(0\) (d) \(-1\) Solution Given, \[ \sin\theta+\cos\theta=1 \] Squaring both sides, \[ (\sin\theta+\cos\theta)^2=1 \] \[ \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1 \] Using […]

If tan A = 1/2 and tan B = 1/3, Find tan(2A + B) If \( \tan A=\frac{1}{2} \) and \( \tan B=\frac{1}{3} \), Find \( \tan(2A+B) \) Question If \[ \tan A=\frac{1}{2},\qquad \tan B=\frac{1}{3}, \] then \[ \tan(2A+B) \] is equal to (a) 1 (b) 2 (c) 3 (d) 4 Solution First find \(\tan

The Value of cos2θ cos2ϕ + sin²(θ−ϕ) − sin²(θ+ϕ) The Value of \( \cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi) \) Question Find the value of \[ \cos2\theta\cos2\phi + \sin^2(\theta-\phi) – \sin^2(\theta+\phi) \] (a) \(\sin2(\theta+\phi)\) (b) \(\cos2(\theta+\phi)\) (c) \(\sin2(\theta-\phi)\) (d) \(\cos2(\theta-\phi)\) Solution Use the identity \[ \sin^2A-\sin^2B = \frac{1-\cos2A}{2} – \frac{1-\cos2B}{2} = \frac{\cos2B-\cos2A}{2} \] Therefore, \[ \sin^2(\theta-\phi)-\sin^2(\theta+\phi) = \frac{\cos2(\theta+\phi)-\cos2(\theta-\phi)}{2} \] Using

The Value of tan75° – cot75° The Value of \( \tan75^\circ-\cot75^\circ \) Question Find the value of \[ \tan75^\circ-\cot75^\circ \] (a) \(2\sqrt3\) (b) \(2+\sqrt3\) (c) \(2-\sqrt3\) (d) \(1\) Solution Use the identity \[ \tan\theta-\cot\theta = \frac{\sin^2\theta-\cos^2\theta} {\sin\theta\cos\theta} \] \[ = \frac{-\cos2\theta} {\frac12\sin2\theta} \] \[ = -2\cot2\theta \] Putting \[ \theta=75^\circ \] \[ \tan75^\circ-\cot75^\circ = -2\cot150^\circ

The Value of (1 – tan²15°)/(1 + tan²15°) The Value of \( \frac{1-\tan^2 15^\circ}{1+\tan^2 15^\circ} \) Question Find the value of \[ \frac{1-\tan^2 15^\circ}{1+\tan^2 15^\circ} \] (a) \(1\) (b) \(\sqrt3\) (c) \(\frac{\sqrt3}{2}\) (d) \(2\) Solution Use the standard identity: \[ \cos 2\theta = \frac{1-\tan^2\theta} {1+\tan^2\theta} \] Putting \[ \theta=15^\circ \] we get \[ \frac{1-\tan^2 15^\circ}

The Value of cos²48° – sin²12° The Value of \( \cos^2 48^\circ-\sin^2 12^\circ \) Question Find the value of \[ \cos^2 48^\circ-\sin^2 12^\circ \] (a) \(\dfrac{\sqrt5+1}{8}\) (b) \(\dfrac{\sqrt5-1}{8}\) (c) \(\dfrac{\sqrt5+1}{5}\) (d) \(\dfrac{\sqrt5+1}{2\sqrt2}\) Solution Use the identities \[ \cos^2\theta=\frac{1+\cos2\theta}{2} \] and \[ \sin^2\theta=\frac{1-\cos2\theta}{2} \] Therefore, \[ \cos^2 48^\circ-\sin^2 12^\circ = \frac{1+\cos96^\circ}{2} – \frac{1-\cos24^\circ}{2} \] \[ =

If tanα = 1/7 and tanβ = 1/3, Find cos2α If \( \tan\alpha=\frac{1}{7} \) and \( \tan\beta=\frac{1}{3} \), Find \( \cos2\alpha \) Question If \[ \tan\alpha=\frac{1}{7},\qquad \tan\beta=\frac{1}{3}, \] then \(\cos2\alpha\) is equal to (a) \(\sin2\beta\) (b) \(\sin4\beta\) (c) \(\sin3\beta\) (d) \(\cos2\beta\) Solution Using the identity \[ \cos2\alpha = \frac{1-\tan^2\alpha} {1+\tan^2\alpha} \] Substituting \(\tan\alpha=\frac{1}{7}\), \[ \cos2\alpha

If tan x = a/b, Find bcos2x + asin2x If \( \tan x=\frac{a}{b} \), Find \( b\cos2x+a\sin2x \) Question If \[ \tan x=\frac{a}{b}, \] then \[ b\cos2x+a\sin2x \] is equal to (a) \(a\) (b) \(b\) (c) \(\frac{a}{b}\) (d) \(\frac{b}{a}\) Solution Using \[ \tan x=\frac{a}{b} \] we have the standard expressions \[ \cos2x=\frac{b^2-a^2}{a^2+b^2} \] and \[

Value of cosα cos2α cos4α … cos2^(n−1)α Value of \( \cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{\,n-1}\alpha \) Question If \(n=1,2,3,\ldots\), then \[ \cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{n-1}\alpha \] is equal to (a) \(\dfrac{\sin(2^n\alpha)}{2^n\sin\alpha}\) (b) \(\dfrac{\sin(2^n\alpha)}{2^n\sin(2^{n-1}\alpha)}\) (c) \(\dfrac{\sin(4^{\,n-1}\alpha)}{4^{\,n-1}\sin\alpha}\) (d) \(\dfrac{\sin(2^n\alpha)}{2^n\sin\alpha}\) Solution Use the standard identity: \[ \sin(2^n\alpha) = 2^n\sin\alpha \cos\alpha \cos2\alpha \cos4\alpha \cdots \cos2^{n-1}\alpha \] Dividing both sides by \(2^n\sin\alpha\), \[ \cos\alpha \cos2\alpha \cos4\alpha \cdots

The Value of sin5x/sinx The Value of \( \frac{\sin 5x}{\sin x} \) Question Find the value of \[ \frac{\sin 5x}{\sin x} \] (a) \(16\cos^4x-12\cos^2x+1\) (b) \(16\cos^4x+12\cos^2x+1\) (c) \(16\cos^4x-12\cos^2x-1\) (d) \(16\cos^4x+12\cos^2x-1\) Solution Use the identity \[ \sin 5x = 16\sin^5x – 20\sin^3x + 5\sin x \] Dividing by \(\sin x\), \[ \frac{\sin 5x}{\sin x} = 16\sin^4x