Ravi Kant Kumar

The Value of cos12° + cos84° + cos156° + cos132° The Value of \( \cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ \) Question Find the value of \[ \cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ \] (a) \(\frac12\) (b) \(1\) (c) \(-\frac12\) (d) \(\frac18\) Solution Use the identity \[ \cos(180^\circ-\theta)=-\cos\theta \] Therefore, \[ \cos156^\circ = -\cos24^\circ \] and \[ \cos132^\circ = -\cos48^\circ \] Hence, \[ \cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ \]

If sinθ = -4/5 and θ Lies in the Third Quadrant, Find cos(θ/2) If \( \sin\theta=-\frac45 \) and \( \theta \) Lies in the Third Quadrant, Find \( \cos\frac{\theta}{2} \) Question If \[ \sin\theta=-\frac45 \] and \(\theta\) lies in the third quadrant, then the value of \[ \cos\frac{\theta}{2} \] is (a) \(\frac15\) (b) \(\frac1{\sqrt{10}}\) (c)

The Value of sin(π/10) sin(13π/10) The Value of \( \sin\frac{\pi}{10}\sin\frac{13\pi}{10} \) Question Find the value of \[ \sin\frac{\pi}{10}\sin\frac{13\pi}{10} \] (a) \(\frac12\) (b) \(-\frac12\) (c) \(-\frac14\) (d) \(1\) Solution Use the identity \[ \sin(\theta+\pi)=-\sin\theta \] Since \[ \frac{13\pi}{10} = \pi+\frac{3\pi}{10}, \] we have \[ \sin\frac{13\pi}{10} = -\sin\frac{3\pi}{10} \] Therefore, \[ \sin\frac{\pi}{10}\sin\frac{13\pi}{10} = -\sin\frac{\pi}{10}\sin\frac{3\pi}{10} \] Now use

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