Evaluate $\tan \left( \frac{1}{2} \cos^{-1} \frac{5}{7} \right)$.

$\frac{1}{\sqrt{6}}$

The correct answer is Option (2) → $\frac{1}{\sqrt{6}}$ ##

Assume that $\frac{1}{2} \cos^{-1} \frac{5}{7} = \theta$

Then, $\cos^{-1} \frac{5}{7} = 2\theta$

$\Rightarrow \cos 2\theta = \frac{5}{7}$

Since, $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$

$∴\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{5}{7}$

$\Rightarrow \tan^2 \theta = \frac{1}{6}$

$\Rightarrow \tan \theta = \left( \sqrt{\frac{1}{6}} \right) \text{ or } \left( -\sqrt{\frac{1}{6}} \right)$

Also, $0 \le \cos^{-1} \frac{5}{7} \le \pi$

$\Rightarrow 0 \le \frac{1}{2} \cos^{-1} \frac{5}{7} \le \frac{\pi}{2}$

$\Rightarrow 0 \le \tan \theta \le \infty$

$∴\tan \theta = \frac{1}{\sqrt{6}}$

[Rejecting the negative value for $\tan \theta$]