Ravi Kant Kumar

Prove that cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8) = 2 Prove that \[ \cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{5\pi}{8}+\cos^2\frac{7\pi}{8}=2 \] Proof: Using the identity \[ \cos(\pi-\theta)=-\cos\theta \] therefore, \[ \cos^2\frac{5\pi}{8}=\cos^2\frac{3\pi}{8} \] and \[ \cos^2\frac{7\pi}{8}=\cos^2\frac{\pi}{8} \] Hence, \[ LHS=2\cos^2\frac{\pi}{8}+2\cos^2\frac{3\pi}{8} \] \[ =2\left(\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}\right) \] Using \[ \cos^2A=\frac{1+\cos2A}{2} \] we get \[ LHS=2\left( \frac{1+\cos\frac{\pi}{4}}{2} + \frac{1+\cos\frac{3\pi}{4}}{2} \right) \] \[ =2\left( \frac{1+\frac{\sqrt2}{2}}{2} +

Prove that cos x/(1 − sin x) = tan(π/4 + x/2) Prove that \[ \frac{\cos x}{1-\sin x}=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right) \] Proof: \[ LHS=\frac{\cos x}{1-\sin x} \] Multiply numerator and denominator by \[ 1+\sin x \] \[ LHS=\frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)} \] Using \[ (1-\sin x)(1+\sin x)=1-\sin^2x \] and \[ 1-\sin^2x=\cos^2x \] we get \[ LHS=\frac{\cos x(1+\sin

Prove that cos 2x/(1 + sin 2x) = tan(π/4 − x) Prove that \[ \frac{\cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right) \] Proof: \[ LHS=\frac{\cos 2x}{1+\sin 2x} \] Using the identities: \[ \cos 2x=\cos^2x-\sin^2x \] \[ \sin 2x=2\sin x\cos x \] Substituting these values: \[ LHS=\frac{\cos^2x-\sin^2x}{1+2\sin x\cos x} \] Factorizing numerator: \[ LHS=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}

Prove that (sin x + sin 2x)/(1 + cos x + cos 2x) = tan x Prove that \[ \frac{\sin x+\sin 2x}{1+\cos x+\cos 2x}=\tan x \] Proof: \[ LHS=\frac{\sin x+\sin 2x}{1+\cos x+\cos 2x} \] Using the identity: \[ \sin 2x=2\sin x\cos x \] and \[ \cos 2x=2\cos^2x-1 \] Substituting these values: \[ LHS=\frac{\sin x+2\sin x\cos

Prove that (1 − cos 2x + sin 2x)/(1 + cos 2x + sin 2x) = tan x Prove that \[ \frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\tan x \] Proof: \[ LHS=\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} \] Using the identities: \[ 1-\cos2x=2\sin^2x \] \[ 1+\cos2x=2\cos^2x \] \[ \sin2x=2\sin x\cos x \] Substituting these values: \[ LHS=\frac{2\sin^2x+2\sin x\cos x}{2\cos^2x+2\sin x\cos x} \] Taking common factors:

Prove that √(2 + √(2 + 2cos 4x)) = 2cos x Prove that \[ \sqrt{2+\sqrt{2+2\cos4x}}=2\cos x \] for \[ 0

Prove that sin 2x /(1 + cos 2x) = tan x Prove that \[ \frac{\sin 2x}{1+\cos 2x}=\tan x \] Proof: \[ LHS=\frac{\sin 2x}{1+\cos 2x} \] Using the identities: \[ \sin 2x=2\sin x\cos x \] \[ 1+\cos 2x=2\cos^2 x \] Substituting these values: \[ LHS=\frac{2\sin x\cos x}{2\cos^2 x} \] \[ =\frac{\sin x\cos x}{\cos^2 x} \] \[

Prove that sin 2x /(1 – cos 2x) = cot x Prove that \[ \frac{\sin 2x}{1-\cos 2x}=\cot x \] Proof: \[ LHS=\frac{\sin 2x}{1-\cos 2x} \] Using the identities: \[ \sin 2x=2\sin x\cos x \] \[ 1-\cos 2x=2\sin^2 x \] Substituting these values: \[ LHS=\frac{2\sin x\cos x}{2\sin^2 x} \] \[ =\frac{\sin x\cos x}{\sin^2 x} \] \[

Prove that √((1 − cos 2x)/(1 + cos 2x)) = tan x Prove that \[ \sqrt{\frac{1-\cos 2x}{1+\cos 2x}}=\tan x \] Proof: \[ LHS=\sqrt{\frac{1-\cos 2x}{1+\cos 2x}} \] Using the identities: \[ 1-\cos 2x=2\sin^2 x \] \[ 1+\cos 2x=2\cos^2 x \] Substituting these values: \[ LHS=\sqrt{\frac{2\sin^2 x}{2\cos^2 x}} \] \[ =\sqrt{\frac{\sin^2 x}{\cos^2 x}} \] \[ =\sqrt{\tan^2 x}

If m sin θ = n sin(θ + 2α), prove that tan(θ + α) cot α = (m + n)/(m − n) If \[ m\sin\theta=n\sin(\theta+2\alpha) \] prove that \[ \tan(\theta+\alpha)\cot\alpha = \frac{m+n}{m-n} \] Solution Given: \[ m\sin\theta=n\sin(\theta+2\alpha) \] Use identity: \[ \sin(A+B)=\sin A\cos B+\cos A\sin B \] \[ m\sin\theta = n[\sin\theta\cos2\alpha+\cos\theta\sin2\alpha] \] Bring terms containing