Question

Prove that ${\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)=\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)$.

Answer 1

${\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)=\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)$

We know that:

${\tan }^{-1}A+{\tan }^{-1}y={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

Therefore,

${\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)={\tan }^{-1}\left(\frac{\frac{2}{9}+\frac{1}{4}}{1-\frac{2\times 1}{9\times 4}}\right)$

$={\tan }^{-1}\left(\frac{\frac{17}{36}}{1-\frac{1}{18}}\right)={\tan }^{-1}\frac{1}{2}$

Now,

${\tan }^{-1}\frac{1}{2}=\frac{1}{2}({\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{2})$

$\Rightarrow {\tan }^{-1}\frac{1}{2}=\frac{1}{2}{\tan }^{-1}\frac{4}{3}=\frac{1}{2}{\sin }^{-1}\frac{4}{5}$