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

The Value of (sin5α – sin3α)/(cos5α + 2cos4α + cos3α) The Value of \( \frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha} \) Question Find the value of \[ \frac{\sin5\alpha-\sin3\alpha} {\cos5\alpha+2\cos4\alpha+\cos3\alpha} \] (a) \(\cot\frac{\alpha}{2}\) (b) \(\cot\alpha\) (c) \(\tan\frac{\alpha}{2}\) (d) none of these Solution Using the identity \[ \sin C-\sin D = 2\cos\frac{C+D}{2} \sin\frac{C-D}{2} \] the numerator becomes \[ \sin5\alpha-\sin3\alpha = 2\cos4\alpha\sin\alpha \]

The Value of tan x + tan(π/3 + x) + tan(2π/3 + x) The Value of \( \tan x+\tan\left(\frac{\pi}{3}+x\right)+\tan\left(\frac{2\pi}{3}+x\right) \) Question Find the value of \[ \tan x+\tan\left(\frac{\pi}{3}+x\right)+\tan\left(\frac{2\pi}{3}+x\right) \] (a) \(3\tan3x\) (b) \(\tan3x\) (c) \(3\cot3x\) (d) \(\cot3x\) Solution Let \[ A=x,\quad B=x+\frac{\pi}{3},\quad C=x+\frac{2\pi}{3} \] Then \[ A+B+C=3x+\pi \] Using the identity \[ \tan(A+B+C) = \frac{\tan

The Value of tan x tan(π/3 – x) tan(π/3 + x) The Value of \( \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right) \) Question Find the value of \[ \tan x\, \tan\left(\frac{\pi}{3}-x\right)\, \tan\left(\frac{\pi}{3}+x\right) \] (a) \(\cot3x\) (b) \(2\cot3x\) (c) \(\tan3x\) (d) \(3\tan3x\) Solution Let \[ t=\tan x \] Using \[ \tan\left(\frac{\pi}{3}-x\right) = \frac{\sqrt3-t}{1+\sqrt3\,t} \] and \[ \tan\left(\frac{\pi}{3}+x\right) = \frac{\sqrt3+t}{1-\sqrt3\,t}

The Value of cos(36° − A)cos(36° + A) + cos(54° − A)cos(54° + A) The Value of \( \cos(36^\circ-A)\cos(36^\circ+A)+\cos(54^\circ-A)\cos(54^\circ+A) \) Question Find the value of \[ \cos(36^\circ-A)\cos(36^\circ+A) + \cos(54^\circ-A)\cos(54^\circ+A) \] (a) \(\cos2A\) (b) \(\sin2A\) (c) \(\cos A\) (d) \(0\) Solution Use the identity \[ \cos(x-y)\cos(x+y) = \cos^2x-\sin^2y \] Therefore, \[ \cos(36^\circ-A)\cos(36^\circ+A) = \cos^236^\circ-\sin^2A \] and

The Value of cos⁴x + sin⁴x – 6cos²x sin²x The Value of \( \cos^4x+\sin^4x-6\cos^2x\sin^2x \) Question Find the value of \[ \cos^4x+\sin^4x-6\cos^2x\sin^2x \] (a) \(\cos2x\) (b) \(\sin2x\) (c) \(\cos4x\) (d) none of these Solution Use the identity \[ \cos^4x+\sin^4x = (\cos^2x+\sin^2x)^2 – 2\cos^2x\sin^2x \] \[ =1-2\cos^2x\sin^2x \] Therefore, \[ \cos^4x+\sin^4x-6\cos^2x\sin^2x = 1-8\cos^2x\sin^2x \] Using \[

If tan x = t, Find tan 2x + sec 2x If \( \tan x=t \), Find \( \tan2x+\sec2x \) Question If \[ \tan x=t, \] then \[ \tan2x+\sec2x \] is equal to (a) \(\dfrac{1+t}{1-t}\) (b) \(\dfrac{1-t}{1+t}\) (c) \(\dfrac{2t}{1-t}\) (d) \(\dfrac{2t}{1+t}\) Solution Using the identities \[ \tan2x=\frac{2t}{1-t^2} \] and \[ \sec2x=\frac{1+t^2}{1-t^2} \] Therefore, \[ \tan2x+\sec2x

If (2^n+1)x=π, Find 2^n cosx cos2x cos2²x … cos2^(n-1)x If \((2^n+1)x=\pi\), Find \(2^n\cos x\cos2x\cos2^2x\cdots\cos2^{n-1}x\) Question If \[ (2^n+1)x=\pi, \] then find \[ 2^n\cos x\cos2x\cos2^2x\cdots\cos2^{n-1}x \] (a) \(-1\) (b) \(1\) (c) \(\frac12\) (d) none of these Solution Use the standard identity: \[ \sin(2^n x) = 2^n \sin x \cos x \cos2x \cos2^2x \cdots \cos2^{n-1}x \] Therefore,

If tan(x/2)=√((1-e)/(1+e)) tan(α/2), Find cosα If \( \tan\frac{x}{2}=\sqrt{\frac{1-e}{1+e}}\tan\frac{\alpha}{2} \), Find \( \cos\alpha \) Question If \[ \tan\frac{x}{2} = \sqrt{\frac{1-e}{1+e}} \tan\frac{\alpha}{2}, \] then \(\cos\alpha\) is equal to (a) \(\dfrac{1-e\cos x}{\cos x+e}\) (b) \(\dfrac{1+e\cos x}{\cos x-e}\) (c) \(\dfrac{1-e\cos x}{\cos x-e}\) (d) \(\dfrac{\cos x-e}{1-e\cos x}\) Solution Let \[ t=\tan\frac{\alpha}{2} \] Then \[ \tan\frac{x}{2} = \sqrt{\frac{1-e}{1+e}}\,t \] Squaring, \[