Prove the identity \(\frac{\cos^2 x - \sin^2 x}{1 + 2 \sin x \cos x} = \frac{1 - \tan x}{1 + \tan x}

Question

Prove the identity $\frac{\cos^2 x - \sin^2 x}{1 + 2 \sin x \cos x} = \frac{1 - \tan x}{1 + \tan x}$.[4] Solve the equation $\frac{\cos^2 x - \sin^2 x}{1 + 2 \sin x \cos x} = \frac{2}{3} \tan x$ for $0 \le x \le 2\pi$.[4]

Accepted Answer

(a) $\frac{\cos^2x-\sin^2x}{1+2\sin x\cos x}=\frac{1-\tan x}{1+\tan x}$ (proven) (b) $x=0.464,\ 1.89,\ 3.61,\ 5.03\text{ rad}$

2025 DSS PRELIMS P2 Q7

2025 DSS PRELIMS P2 Q7

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