Question

Find the value of ${\tan }^2\left(\frac{1}{2}{\sin }^{-1}\frac{2}{3}\right)$

Answer 1

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

$\Rightarrow \cos 2\theta +\cos 2\theta {\tan }^2\theta =1-{\tan }^2\theta$

$\Rightarrow {\tan }^2\theta (\cos 2\theta +1)=1-\cos 2\theta$

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

Now, ${\tan }^2\left(\frac{1}{2}{\sin }^{-1}\frac{2}{3}\right)=\frac{1-\cos (2\times \frac{1}{2}{\sin }^{-1}\frac{2}{3})}{1+\cos (2\times \frac{1}{2}{\sin }^{-1}\frac{2}{3})}$

$=\frac{1-\cos \left({\sin }^{-1}\frac{2}{3}\right)}{1+\cos \left({\sin }^{-1}\frac{2}{3}\right)}$

Let ${\sin }^{-1}\frac{2}{3}=m$

$\Rightarrow \frac{2}{3}=\sin m$

$\Rightarrow \cos m=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$

$\Rightarrow m={\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)$

So, by using the value we get

$=\frac{1-\cos \left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)}{1+\cos \left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)}$

$=\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}=\frac{3-\sqrt{5}}{3+\sqrt{5}}$.