Show that ^ -1 5 13 + ^ -1 3 5 = ^ -1 63 16 . - Vidyadip

Question

Show that $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}=\tan^{-1}\frac{63}{16}.$

✓

Answer

Let us suppose, $\sin^{-1}\frac{5}{13}=\text{x},\ \cos^{-1}\frac{3}{5}=\text{y}$ $\Rightarrow\ \sin\text{x}=\frac{5}{13}$ and $\cos\text{y}=\frac{3}{5}$$\Rightarrow\ \cos\text{x}=\sqrt{1-\Big(\frac{5}{13}\Big)^2}=\frac{12}{13}$ and $\sin\text{y}=\sqrt{1-\Big(\frac{3}{5}\Big)^2}=\frac{4}{5}$
Now, $\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$ $\Rightarrow\ \tan\text{x}=\frac{5}{12}$ $\Rightarrow\ \text{x}=\tan^{-1}\frac{5}{12}$ And $\tan\text{y}=\frac{\sin\text{y}}{\cos\text{y}}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$ $\Rightarrow\ \text{y}=\tan^{-1}\frac{4}{3}$ Now, LHS $=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$ $=\text{x}+\text{y}$ $=\tan^{-1}\text{x}+\tan^{-1}\text{y}$ $=\tan^{-1}\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}$ $=\tan^{-1}\Bigg(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}\Bigg)$ $=\tan^{-1}\frac{63}{16}$ = RHS

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