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

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