Prove that ${\sin ^{ – 1}}\frac{8}{{17}} + {\sin ^{ – 1}}\frac{3}{5} = {\sin ^{ – 1}}\frac{{77}}{{85}}$.
[NCERT,Exemplar.2.3,Q.14,Page.36]

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.


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