Question

Prove that:
\( \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right), \quad x \in [0, 1] \)

Answer 1

Let us assume: \[ \theta = \tan^{-1}(\sqrt{x}) \Rightarrow \tan \theta = \sqrt{x} \] where \( \theta \in \left[0, \frac{\pi}{2}\right] \)

Squaring both sides: \[ \tan^2 \theta = x \Rightarrow x = \tan^2 \theta \] Now use the identity: \[ \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - x}{1 + x} \] Taking inverse cosine: \[ \cos^{-1}\left( \frac{1 - x}{1 + x} \right) = \cos^{-1}(\cos(2\theta)) = 2\theta \Rightarrow \theta = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right) \]

Since \( \theta = \tan^{-1}(\sqrt{x}) \), we conclude: \[ \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right) \quad \text{for } x \in [0, 1] \]

Final Result:

\[ \boxed{ \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right) } \]