Evaluate : $\sin^{-1}\large\frac{4}{5}$$+\sin^{-1}\large\frac{5}{13}$$+\sin^{-1}\large\frac{16}{65}$

$(a)\;1\qquad(b)\;\large\frac{\pi}{2}\qquad$$(c)\;\large\frac{\pi}{4}$$\qquad(d)\;0$

$\sin^{-1}\big(\large\frac{4}{5}$$.\sqrt{1-\large\frac{25}{169}}+\sqrt{1-\large\frac{16}{25}}.\large\frac{5}{13}\big)$$+\sin^{-1}\large\frac{16}{65}$
$\Rightarrow \sin^{-1}\big(\large\frac{48}{65}+\frac{15}{65}\big)$$+\sin^{-1}\large\frac{16}{65} \Rightarrow \sin^{-1}\large\frac{63}{65}$$+\sin^{-1}\large\frac{16}{65}$
$\Rightarrow \sin^{-1}\big(\large\frac{63}{65}\sqrt{1-\big(\large\frac{16}{65}}\big)^2+\frac{16}{65}\sqrt{1-\big(\large\frac{63}{65}}\big)^2\big)$
$\Rightarrow \sin^{-1}\big(\large\frac{63}{65}.\frac{63}{65}+\frac{16}{65}.\frac{16}{65}\big)$
$\Rightarrow \sin^{-1}\big(\large\frac{63^2+16^2}{65^2}\big)$
$\Rightarrow \sin^{-1}\big(\large\frac{65^2}{65^2}\big)$
$\Rightarrow \sin^{-1}1$
$\Rightarrow \large\frac{\pi}{2}$
Hence (b) is the correct answer.