Solve 2tan⁻¹(sin x) = tan⁻¹(2sec x)

Solve \( 2\tan^{-1}(\sin x) = \tan^{-1}(2\sec x), \quad x \ne \frac{\pi}{2} \)

Solution:

Use identity:

\[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \]

Let

\[ \theta = \tan^{-1}(\sin x) \Rightarrow \tan \theta = \sin x \]

Then,

\[ \tan(2\theta) = \frac{2\sin x}{1 – \sin^2 x} = \frac{2\sin x}{\cos^2 x} = 2\sin x \sec^2 x \]

Given:

\[ 2\tan^{-1}(\sin x) = \tan^{-1}(2\sec x) \Rightarrow \tan(2\theta) = 2\sec x \]

So,

\[ 2\sin x \sec^2 x = 2\sec x \]

\[ \Rightarrow \sin x \sec x = 1 \]

\[ \Rightarrow \frac{\sin x}{\cos x} = 1 \Rightarrow \tan x = 1 \]

\[ x = \frac{\pi}{4} + n\pi \]

Given restriction \(x \ne \frac{\pi}{2}\) is satisfied.

\[ x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]

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