If cos2α = (3cos2β – 1)/(3 – cos2β), Find tanα If \( \cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta} \), Find \( \tan\alpha \) Question If \(\alpha\) and \(\beta\) are acute angles satisfying \[ \cos2\alpha = \frac{3\cos2\beta-1} {3-\cos2\beta}, \] then \(\tan\alpha\) is equal to (a) \(\sqrt2\tan\beta\) (b) \(\frac1{\sqrt2}\tan\beta\) (c) \(\sqrt2\cot\beta\) (d) \(\frac1{\sqrt2}\cot\beta\) Solution Use the identity \[ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta} \] Let \[ […]

2(1 – 2sin²7x)sin3x is Equal to What? \(2(1-2\sin^2 7x)\sin 3x\) is Equal to What? Question Find the value of \[ 2(1-2\sin^2 7x)\sin 3x \] (a) \(\sin17x-\sin11x\) (b) \(\sin11x-\sin17x\) (c) \(\cos17x-\cos11x\) (d) \(\cos17x+\cos11x\) Solution Using the identity \[ 1-2\sin^2\theta=\cos2\theta \] we get \[ 2(1-2\sin^2 7x)\sin3x = 2\cos14x\sin3x \] Now use \[ 2\sin A\cos B = \sin(A+B)+\sin(A-B)

The Value of 2(sin2x + 2cos²x – 1)/(cosx – sinx – cos3x + sin3x) The Value of \( \frac{2(\sin2x+2\cos^2x-1)}{\cos x-\sin x-\cos3x+\sin3x} \) Question Find the value of \[ \frac{2(\sin2x+2\cos^2x-1)} {\cos x-\sin x-\cos3x+\sin3x} \] (a) \(\cos x\) (b) \(\sec x\) (c) \(\csc x\) (d) \(\sin x\) Solution First simplify the numerator: \[ 2\cos^2x-1=\cos2x \] Therefore, \[

The Value of 2sin²B + 4cos(A+B)sinA sinB + cos2(A+B) The Value of \(2\sin^2B + 4\cos(A+B)\sin A\sin B + \cos2(A+B)\) Question Find the value of \[ 2\sin^2B + 4\cos(A+B)\sin A\sin B + \cos2(A+B) \] (a) \(0\) (b) \(\cos3A\) (c) \(\cos2A\) (d) none of these Solution Use the identity \[ 2\sin^2B = 1-\cos2B \] and \[ \cos2(A+B)=2\cos^2(A+B)-1

The Value of sin3x/(1 + 2cos2x) is Equal to What? The Value of \( \frac{\sin 3x}{1+2\cos 2x} \) is Equal to What? Question Find the value of \[ \frac{\sin 3x}{1+2\cos 2x} \] (a) \(\cos x\) (b) \(\sin x\) (c) \(-\cos x\) (d) \(\sin x\) Solution Use the identity \[ \sin 3x=\sin x+2\sin x\cos 2x \]

The Value of cos²(π/6 + x) – sin²(π/6 – x) The Value of \( \cos^2\left(\frac{\pi}{6}+x\right)-\sin^2\left(\frac{\pi}{6}-x\right) \) Question Find the value of \[ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) \] (a) \(\frac{1}{2}\cos2x\) (b) \(0\) (c) \(-\frac{1}{2}\cos2x\) (d) \(\frac{1}{2}\) Solution Use the identities \[ \cos^2\theta=\frac{1+\cos2\theta}{2} \] and \[ \sin^2\theta=\frac{1-\cos2\theta}{2} \] Therefore, \[ \cos^2\left(\frac{\pi}{6}+x\right) = \frac{1+\cos\left(\frac{\pi}{3}+2x\right)}{2} \] \[ \sin^2\left(\frac{\pi}{6}-x\right) = \frac{1-\cos\left(\frac{\pi}{3}-2x\right)}{2}

If tan(π/4 + x) + tan(π/4 – x) = λ sec2x, Find λ If \( \tan\left(\frac{\pi}{4}+x\right)+\tan\left(\frac{\pi}{4}-x\right)=\lambda\sec2x \), Find \( \lambda \) Question If \[ \tan\left(\frac{\pi}{4}+x\right) + \tan\left(\frac{\pi}{4}-x\right) = \lambda\sec2x, \] then \(\lambda\) is (a) 3 (b) 4 (c) 1 (d) 2 Solution Let \[ A=\frac{\pi}{4}+x,\qquad B=\frac{\pi}{4}-x \] Using the identity \[ \tan A+\tan B =

The Value of cos3x/(2cos2x – 1) is Equal to What? The Value of \( \frac{\cos 3x}{2\cos 2x-1} \) is Equal to What? Question Find the value of \[ \frac{\cos 3x}{2\cos 2x-1} \] (a) \(\cos x\) (b) \(\sin x\) (c) \(\tan x\) (d) none of these Solution Use the identity \[ \cos 3x = 4\cos^3x –

If A = 2sin²x – cos2x, Find the Interval in Which A Lies If \(A = 2\sin^2x – \cos2x\), Find the Interval in Which A Lies Question If \[ A = 2\sin^2x – \cos2x, \] then \(A\) lies in the interval (a) \([-1,3]\) (b) \([1,2]\) (c) \([-2,4]\) (d) none of these Solution Using the identity

The Value of 2cosx – cos3x – cos5x – 16cos³x sin²x The Value of \(2\cos x-\cos3x-\cos5x-16\cos^3x\sin^2x\) Question Find the value of \[ 2\cos x-\cos3x-\cos5x-16\cos^3x\sin^2x \] (a) \(2\) (b) \(1\) (c) \(0\) (d) \(-1\) Solution Use the identity: \[ \cos3x+\cos5x = 2\cos4x\cos x \] Therefore, \[ 2\cos x-\cos3x-\cos5x = 2\cos x-2\cos4x\cos x \] \[ = 2\cos