May 2026

If π/2 < x < π, Then Find the Value of √[(1 − cos 2x)/(1 + cos 2x)] Question If \[ \frac{\pi}{2}

If π/2 < x < π, Then Find the Simplest Form of √(2 + √(2 + 2cos2x)) Question If \[ \frac{\pi}{2}

If π/2 < x < 3π/2, Then Find √[(1 + cos 2x)/2] Question If \[ \frac{\pi}{2}

If tan(x/2) = m/n, Then Find m sin x + n cos x Question: \[ \tan\frac{x}{2}=\frac{m}{n} \] Find \[ m\sin x+n\cos x. \] Solution Using the identities \[ \sin x=\frac{2\tan(x/2)}{1+\tan^2(x/2)} \] and \[ \cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} \] Since \[ \tan\frac{x}{2}=\frac{m}{n}, \] we get \[ \sin x = \frac{2(m/n)}{1+m^2/n^2} = \frac{2mn}{m^2+n^2} \] and \[ \cos x =

Value of Trigonometric Functions at Multiples and Submultiples of an angle – Very Shorts Answer Questions (VSAQs) Exercise Solutions If cos 4x = 1 + k sin² x cos² x, then write the value of k. Watch Solution If tan x/2 = m/n, then write the value of m sin x + n cos x. Watch

If cos 4x = 1 + k sin²x cos²x, Then Find k Question: \[ \cos 4x=1+k\sin^2x\cos^2x \] Find the value of \(k\). Solution Using the identity \[ \cos 4x = 1-2\sin^2 2x \] Also, \[ \sin 2x=2\sin x\cos x \] \[ \sin^2 2x = 4\sin^2 x\cos^2 x \] Substituting, \[ \cos 4x = 1-2(4\sin^2 x\cos^2

Find the Value of 108 sin(π/9) − 144 sin(3π/9) Question: \[ 108\sin\frac{\pi}{9}-144\sin\frac{3\pi}{9} \] Solution Since \[ \frac{3\pi}{9}=\frac{\pi}{3} \] the expression becomes \[ 108\sin20^\circ-144\sin60^\circ \] \[ =108\sin20^\circ-144\left(\frac{\sqrt3}{2}\right) \] \[ =108\sin20^\circ-72\sqrt3 \] Now use the standard identity \[ \sin3\theta = 3\sin\theta-4\sin^3\theta \] For \(\theta=20^\circ\), \[ \sin60^\circ = 3\sin20^\circ-4\sin^320^\circ \] \[ \frac{\sqrt3}{2} = 3\sin20^\circ-4\sin^320^\circ \] The known exact

If π/4 < x < π/2, Then Find √(2 + √(2 + 2cos4x)) Question: \[ \frac{\pi}{4}

If π/2 < x < π, Then Find √[(1 − cos 2x)/(1 + cos 2x)] Question: \[ \frac{\pi}{2}

Find the Value of cos²6° – cos²24° Question: \[ \cos^2 6^\circ-\cos^2 24^\circ \] Solution Use the identity \[ \cos^2 A-\cos^2 B = (\cos A-\cos B)(\cos A+\cos B) \] Also, \[ \cos C-\cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \] \[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \] Therefore, \[ \cos^2 A-\cos^2 B = -\sin(C+D)\sin(C-D) \] Taking \(A=6^\circ\) and \(B=24^\circ\),