Inverse Trigonometric Functions — Class 12 Maths Solution

We have, ${\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\frac{{77}}{{85}}$

$therefore,$ ${\rm{LHS}} = {\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5}$

$= {\tan ^{ - 1}}\frac{8}{{15}} + {\tan ^{ - 1}}\frac{3}{4}$

Let ${\sin ^{ - 1}}\frac{8}{{17}} = {\theta _1} \Rightarrow \sin {\theta _1} = \frac{8}{{17}}$

$\Rightarrow$ $\tan {\theta _1} = \frac{8}{{15}} \Rightarrow {\theta _1} = {\tan ^{ - 1}}\frac{8}{{15}}$

and ${\sin ^{ - 1}}\frac{3}{5} = {\theta _2} \Rightarrow \sin {\theta _2} = \frac{3}{5}$

$\Rightarrow$ $\tan {\theta _2} = \frac{3}{4} \Rightarrow {\theta _2} = {\tan ^{ - 1}}\frac{3}{4}$

$= {\tan ^{ - 1}}\left[ {\frac{{\frac{8}{{15}} + \frac{3}{4}}}{{1 - \frac{8}{{15}} \times \frac{3}{4}}}} \right]$

$= {\tan ^{ - 1}}\left[ {\frac{{\frac{{32 + 45}}{{60}}}}{{\frac{{60 - 24}}{{60}}}}} \right] = {\tan ^{ - 1}}\left( {\frac{{77}}{{36}}} \right)$

Let ${\theta _3} = {\tan ^{ - 1}}\frac{{77}}{{36}} \Rightarrow \tan {\theta _3} = \frac{{77}}{{36}}$

$\Rightarrow$ $\sin {\theta _3} = \frac{{77}}{{\sqrt {5929 + 1296} }} = \frac{{77}}{{85}}$

$therefore,$ ${\theta _3} = {\sin ^{ - 1}}\frac{{77}}{{85}}$

$= {\sin ^{ - 1}}\frac{{77}}{{85}} = {\rm{RHS}}$
Hence proved.

Alternate Method To prove, ${\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\frac{{77}}{{85}}$

Let ${\sin ^{ - 1}}\frac{8}{{17}} = x$
$\Rightarrow$ $\sin x = \frac{8}{{17}}$
$\Rightarrow$ $\cos x = \sqrt {1 - {{\sin }^2}x} = \sqrt {1 - {{\left( {\frac{8}{{17}}} \right)}^2}}$

$= \sqrt {\frac{{289 - 64}}{{289}}} = \sqrt {\frac{{225}}{{289}}} = \frac{{15}}{{17}}$

Let ${\sin ^{ - 1}}\frac{3}{5} = y$
$\Rightarrow$ $\sin y = \frac{3}{5} \Rightarrow {\sin ^2}y = \frac{9}{{25}}$

$therefore,$ ${\cos ^2}y = 1 - \frac{9}{{25}}$

$\Rightarrow$ ${\cos ^2}y = {\left( {\frac{4}{5}} \right)^2} \Rightarrow \cos y = \frac{4}{5}$

Now, $\sin (x + y) = \sin x \cdot \cos y + \cos x \cdot \sin y$

$= \frac{8}{{17}} \cdot \frac{4}{5} + \frac{{15}}{{17}} \cdot \frac{3}{5}$
$= \frac{{32}}{{85}} + \frac{{45}}{{85}} = \frac{{77}}{{85}}$

$\Rightarrow$ $(x + y) = {\sin ^{ - 1}}\left( {\frac{{77}}{{85}}} \right)$
$\Rightarrow$ ${\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\frac{{77}}{{85}}$