Question

Solve for $x: 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$.

Answer 1

SOLUTION : We have $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$

$\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right]$

$\therefore \quad \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \Rightarrow \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow 2 \sin ^2 x=2 \sin x \cos x$

$2 \sin ^2 x-2 \sin x \cos x=0 \Rightarrow 2 \sin x(\sin x-\cos x)=0$

$\therefore \quad \sin x=0$ or $\sin x-\cos x=0 \Rightarrow \sin x=\sin 0$ or $\sin x=\cos x$

$\Rightarrow \quad \mathrm{x}=0$ or $\tan \mathrm{x}=1=\tan \frac{\pi}{4} \quad \therefore \quad \mathrm{x}=0$ or $\mathrm{x}=\frac{\pi}{4}$