Ravi Kant Kumar

If cos⁶x + sin⁶x + k sin²2x = 1, Then Find k Question: \[ \cos^6 x+\sin^6 x+k\sin^2 2x=1 \] Find the value of \(k\). Solution Use the identity \[ a^3+b^3=(a+b)^3-3ab(a+b) \] Let \[ a=\cos^2 x,\qquad b=\sin^2 x \] Then \[ \cos^6 x+\sin^6 x =(\cos^2 x+\sin^2 x)^3 -3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x) \] \[ =1-3\sin^2 x\cos^2

Find the Least Value of 2 sin²θ + 3 cos²θ Question: \[ 2\sin^2\theta+3\cos^2\theta \] Find its least value. Solution Using the identity \[ \sin^2\theta+\cos^2\theta=1 \] Write \[ 2\sin^2\theta+3\cos^2\theta \] \[ =2(\sin^2\theta+\cos^2\theta)+\cos^2\theta \] \[ =2+\cos^2\theta \] Since \[ 0\le \cos^2\theta \le 1 \] the minimum value of \(\cos^2\theta\) is \(0\). Therefore, \[ 2+\cos^2\theta \ge 2 \]

Find the Value of cos²48° – sin²12° Question: \[ \cos^2 48^\circ-\sin^2 12^\circ \] Solution Use the identity \[ \sin^2\theta=\cos^2(90^\circ-\theta) \] Therefore, \[ \sin^2 12^\circ=\cos^2 78^\circ \] Hence, \[ \cos^2 48^\circ-\sin^2 12^\circ = \cos^2 48^\circ-\cos^2 78^\circ \] Using \[ \cos^2 A=\frac{1+\cos 2A}{2} \] \[ =\frac{1+\cos96^\circ}{2} -\frac{1+\cos156^\circ}{2} \] \[ =\frac{\cos96^\circ-\cos156^\circ}{2} \] Using the identity \[ \cos C-\cos

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