Show that ${\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{63}}{{16}}$.
[NCERT,Exemplar.2.3,Q.15,Page.36]

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

…….(i)
Let ${\sin ^{ - 1}}\frac{5}{{13}} = x$
$ \Rightarrow $$\sin x = \frac{5}{{13}}$
and ${\cos ^2}x = 1 - {\sin ^2}x$

$ = 1 - \frac{{25}}{{169}} = \frac{{144}}{{169}}$
$ \Rightarrow $$\cos x = \sqrt {\frac{{144}}{{169}}} = \frac{{12}}{{13}}$

$therefore, $$\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{5/13}}{{12/13}} = \frac{5}{{12}}$

……(ii)
$ \Rightarrow $$\tan x = 5/12$

…….(iii)
Again, let ${\cos ^{ - 1}}\frac{3}{5} = y \Rightarrow \cos y = \frac{3}{5}$

$therefore, $$\sin y = \sqrt {1 - {{\cos }^2}y} $
$ = \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {1 - \frac{9}{{25}}} $

$\sin y = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
$ \Rightarrow $$\tan y = \frac{{\sin y}}{{\cos y}} = \frac{{4/5}}{{3/5}} = \frac{4}{3}$

…….(iii)
We know that,
$\tan (x + y) = \frac{{\tan x + \tan y}}{{1 - \tan x \cdot \tan y}}$

$ \Rightarrow $$\tan (x + y) = \frac{{\frac{5}{{12}} + \frac{4}{3}}}{{1 - \frac{5}{{12}} \cdot \frac{4}{3}}} \Rightarrow \tan (x + y) = \frac{{\frac{{15 + 48}}{{36}}}}{{\frac{{36 - 20}}{{36}}}}$

$ \Rightarrow $$\tan (x + y) = \frac{{63/36}}{{16/36}}$

$ \Rightarrow $$\tan (x + y) = \frac{{63}}{{16}}$
$ \Rightarrow $$x + y = {\tan ^{ - 1}}\frac{{63}}{{16}}$

$ \Rightarrow $${\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{4}{3} = {\tan ^{ - 1}}\frac{{63}}{{16}}$ Hence proved

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