Prove that: `sin^(-1)((63)/(65))=sin^(-1)(5/(13))+cos^(-1)(3/5)`

Answer

proved

Explanation

\begin{aligned}
&\mathrm{RHS}=\sin ^{-1}(5) /(13)+\cos ^{\wedge}(-1) \frac{3}{5}\\
&=\tan ^{-1} \frac{\frac{5}{13}}{\sqrt{1-\left(\frac{5}{13}\right)^{2}}}+\tan ^{-1} \frac{\sqrt{1-\left(\frac{3}{5}\right)^{2}}}{\frac{3}{5}}\\ \\
&(\left.\therefore \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} \text { and } \cos ^{-1} x=\tan ^{-1} \frac{\sqrt{1-x^{2}}}{x}\right)\\ \\
&=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3}\\ \\
&=\tan ^{-1} \frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}=\tan ^{-1} \frac{\frac{15+48}{36}}{\frac{36-20}{36}}\\ \\
&\tan ^{-1} \frac{63}{16}=\text { LHS } \\ \\
& \text { Hence Proved. }
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