Inverse Trigonometric Functions — Class 12 Maths Solution

Let $x = {\cos ^{ - 1}}\frac{4}{5}$ and $y = {\cos ^{ - 1}}\frac{{12}}{{13}}$
$\Rightarrow$ $\cos x = \frac{4}{5}\;\;and\;\;\cos y = \frac{{12}}{{13}}$

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

$\Rightarrow$ $\sin x = \sqrt {1 - \frac{{16}}{{25}}} \;\;and\;\;\sin y = \sqrt {1 - \frac{{144}}{{169}}}$
$\Rightarrow$ $\sin x = \frac{3}{5}\;\;and\;\;\sin y = \frac{5}{{13}}$

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

$\Rightarrow$ $\cos (x + y) = \frac{{48}}{{65}} - \frac{{15}}{{65}} = \frac{{33}}{{65}} \Rightarrow x + y = {\cos ^{ - 1}}\left( {\frac{{33}}{{65}}} \right)$

Or, ${\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{{12}}{{13}} = {\cos ^{ - 1}}\left( {\frac{{33}}{{65}}} \right)$