Question

If $y={\cot }^{-1}\left(\sqrt{\cos \theta }\right)-{\tan }^{-1}\sqrt{\cos \theta }$. Prove that $\sin y={\tan }^2\frac{\theta }{2}$.

Answer 1

$y=co{t}^{-1}\sqrt{cos\theta }-ta{n}^{-1}\left(\sqrt{cos\theta }\right)$

Let $\sqrt{cos\theta }=x$

$y=co{t}^{-1}x-ta{n}^{-1}x$

$=ta{n}^{-1}\frac{1}{x}-ta{n}^{-1}x$

$=ta{n}^{-1}\left(\frac{\frac{1}{x}-x}{1+1}\right)$

$=ta{n}^{-1}\left(\frac{1-x^2}{2x}\right)$

$=si{n}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$=si{n}^{-1}\left(\frac{1-cos\theta }{1+cos\theta }\right)$

$=si{n}^{-1}\left(ta{n}^2\theta /2\right)$

$siny=ta{n}^2\theta /2$