Question

Prove that ${\tan }^{-1}\frac{63}{16}={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}$

Answer 1

${\tan }^{-1}\frac{63}{16}={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}$

We can write:

${\sin }^{-1}\frac{5}{13}={\tan }^{-1}\frac{5}{12}$

${\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{4}{3}$

Now,

We know that:

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

Therefore,

${\tan }^{-1}\left(\frac{5}{12}\right)+{\tan }^{-1}\left(\frac{4}{3}\right)={\tan }^{-1}\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5\times 4}{12\times 3}}\right)$

$={\tan }^{-1}\left(\frac{\frac{21}{12}}{1-\frac{5}{9}}\right)={\tan }^{-1}\frac{63}{16}$