Inverse Trigonometric Functions — Class 12 Maths Solution

Let $x = {\cos ^{ - 1}}\frac{{12}}{{13}}\;\;and\;\;y = {\sin ^{ - 1}}\frac{3}{5}$
Or $\cos x = \frac{{12}}{{13}}\;\;and\;\;\sin y = \frac{3}{5}$

Now, $\sin x = \sqrt {1 - {{\cos }^2}x} \;\;and\;\;\cos y = \sqrt {1 - {{\sin }^2}y}$

$\Rightarrow$ $\sin x = \sqrt {1 - \frac{{144}}{{169}}} \;\;and\;\;\cos y = \sqrt {1 - \frac{9}{{25}}}$
$\Rightarrow$ $\sin x = \frac{5}{{13}}\;\;and\;\;\cos y = \frac{4}{5}$

As we know that, $\sin (x + y) = \sin x\cos y + \cos x\sin y$
$= \frac{5}{{13}} \times \frac{4}{5} + \frac{{12}}{{13}} \times \frac{3}{5} = \frac{{20}}{{65}} + \frac{{36}}{{65}} = \frac{{56}}{{65}}$

$\Rightarrow$ $x + y = {\sin ^{ - 1}}\left( {\frac{{56}}{{65}}} \right)\;\;or,\;\;{\cos ^{ - 1}}\left( {\frac{{12}}{{13}}} \right) + {\sin ^{ - 1}}\left( {\frac{3}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{56}}{{65}}} \right)$