May 2026

In a Triangle ABC with ∠C = π/2, Find the Equation Whose Roots Are tan A and tan B Question: In a triangle ABC with \[ \angle C=\frac{\pi}{2}, \] find the equation whose roots are \[ \tan A \quad \text{and} \quad \tan B. \] Solution Since \[ A+B=\frac{\pi}{2} \] we have \[ \tan(A+B)=\tan\frac{\pi}{2} \] Hence, […]

If k = sin(π/18) sin(5π/18) sin(7π/18), Then Find the Value of k Question: \[ k=\sin\frac{\pi}{18}\sin\frac{5\pi}{18}\sin\frac{7\pi}{18} \] Find the numerical value of \(k\). Solution Convert the angles into degrees: \[ k=\sin10^\circ\sin50^\circ\sin70^\circ \] Use the standard identity \[ \sin10^\circ\sin50^\circ\sin70^\circ=\frac{1}{8} \] Therefore, \[ k=\frac{1}{8} \] Verification \[ \sin50^\circ=\cos40^\circ,\qquad \sin70^\circ=\cos20^\circ \] \[ k=\sin10^\circ\cos20^\circ\cos40^\circ \] Using the identity \[ \sin

If tan x = (1 – cos y)/sin y, Then Find tan 2x Question: \[ \tan x=\frac{1-\cos y}{\sin y} \] Find the value of \(\tan 2x\). Solution Using the standard half-angle identity \[ \tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} \] Comparing with the given expression, \[ \tan x = \frac{1-\cos y}{\sin y} = \tan\frac{y}{2} \] Therefore, \[ x=\frac{y}{2}

Find the Value of cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15) Question: \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} \] Find its value. Solution Use the identity \[ \cos x \cos 2x \cos 4x \cos 8x = \frac{\sin 16x}{16\sin x} \] Let \[ x=\frac{2\pi}{15} \] Then \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} = \frac{\sin\left(\frac{32\pi}{15}\right)} {16\sin\left(\frac{2\pi}{15}\right)} \] Now, \[ \sin\frac{32\pi}{15} = \sin\left(2\pi+\frac{2\pi}{15}\right)

If cos x cos 2x cos 2²x … cos 2ⁿ⁻¹x = λ sin(2ⁿx)/sin x, Then Find λ Question: \[ \cos x \cos 2x \cos 2^2x \cdots \cos 2^{n-1}x = \lambda \frac{\sin(2^n x)}{\sin x} \] Find the value of \(\lambda\). Solution Use the standard identity: \[ \sin 2A = 2\sin A\cos A \] Therefore, \[ \sin

If {1 – tan²(π/4 – x)}/{1 + tan²(π/4 – x)} = sin kx, Then Find k Question: \[ \frac{1-\tan^2\left(\frac{\pi}{4}-x\right)} {1+\tan^2\left(\frac{\pi}{4}-x\right)} = \sin kx \] Then find the value of \(k\). Solution Using the identity \[ \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta \] Let \[ \theta=\frac{\pi}{4}-x \] Therefore, \[ \frac{1-\tan^2\left(\frac{\pi}{4}-x\right)} {1+\tan^2\left(\frac{\pi}{4}-x\right)} = \cos\left[2\left(\frac{\pi}{4}-x\right)\right] \] \[ =\cos\left(\frac{\pi}{2}-2x\right) \] Using

Value of Trigonometric Functions at Multiples and Submultiples of an angle – Fill in the Blanks Type Questions (FBQs) Exercise Solutions If {1 – tan² ( π/4 – x )}/{1 + tan² ( π/4 – x )} = sin kx, then k = ………. Watch Solution If cos x cos 2x cos 2² x ……….. cos

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