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# Prove that : $\sin^{-1} \bigg(\large \frac{4}{5} \bigg)$$+\sin^{-1} \bigg(\large \frac{5}{13} \bigg)$$+ \sin^{-1} \bigg( \large\frac{16}{65} \bigg) = \large\frac{\pi}{2}$

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• $\sin^{-1} x+\sin^{-1}y=\sin^{-1}\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\}$
Given :$\sin^{-1}\bigg(\large\frac{4}{5}\bigg)$$+\sin^{-1}\bigg(\large\frac{5}{13}\bigg)$$+\sin^{-1}\bigg(\large\frac{16}{65}\bigg)$
$\Rightarrow \sin^{-1}\big[\large\frac{4}{5}$$\sqrt{1-(\large\frac{5}{13})^2}+\large\frac{5}{13}\sqrt{1-(\large\frac{4}{5})^2}\bigg]+\sin^{-1}\large\frac{16}{65} \Rightarrow \sin^{-1}\bigg[\large\frac{4}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{3}{5}\bigg]+\sin^{-1}\big(\large\frac{16}{25}\big) \Rightarrow \sin^{-1}\large\frac{63}{65}$$+\sin^{-1}\large\frac{16}{65}$
$\sin^{-1}\large\frac{63}{65}\big)$$=\cos^{-1}\sqrt{1-(\large\frac{63}{65})^2}=\cos^{-1}\large\frac{16}{65} \Rightarrow \cos^{-1}\large\frac{16}{65}$$+\sin^{-1}\large\frac{16}{65}$
But $\sin^{-1}x+\cos^{-1}x=\large\frac{\pi}{2}$
$\Rightarrow \large\frac{\pi}{2}$