Question

Prove that:
${\sin }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{1}{7}=\frac{\pi }{4}$

Answer 1

Now,

${\sin }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{1}{7}$

$={\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{1}{7}$ [ Since ${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{\sqrt{5^2-3^2}}={\tan }^{-1}\frac{3}{4}$]

$={\tan }^{-1}\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\times \frac{1}{7}}$

$={\tan }^{-1}\frac{\frac{25}{28}}{\frac{25}{28}}$

$={\tan }^{-1}\frac{25}{25}$

$={\tan }^{-1}1$

$=\frac{\pi }{4}$.