Question

Write the following into simple test form;
(1) $\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-n}{1+n}}\right\}$

Answer 1

Given an expression:

$\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-n}{1+n}}\right\}$

Substitute $n=\cos 2\theta$ in the above expression

$\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right\}=\sin \left\{2{\tan }^{-1}\sqrt{\frac{2\sin {}^2\theta }{2\cos {}^2\theta }}\right\}=\sin \left\{2{\tan }^{-1}\tan \theta \right\}=\sin \left\{2\theta \right\}$

But we have $n=\cos 2\theta$

$\therefore \sin \left\{2{\tan }^{-1}\sqrt{\frac{1-n}{1+n}}\right\}=\sin 2\theta =\sqrt{1-\cos {}^{2}2\theta }=\sqrt{1-n^2}$