Prove that:

To Prove: $\sin ^{-1} \frac{1}{\sqrt{5}}+\sin ^{-1} \frac{2}{\sqrt{5}}=\frac{\pi}{2}$

Formula Used: $\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \times \sqrt{1-y^{2}}+y \times \sqrt{1-x^{2}}\right)$

Proof:

$\mathrm{LHS}=\sin ^{-1} \frac{1}{\sqrt{5}}+\sin ^{-1} \frac{2}{\sqrt{5}}$

$=\sin ^{-1}\left(\frac{1}{\sqrt{5}} \times \sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^{2}}+\frac{2}{\sqrt{5}} \times \sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^{2}}\right)$

$=\sin ^{-1}\left(\frac{1}{\sqrt{5}} \times \sqrt{1-\frac{4}{5}}+\frac{2}{\sqrt{5}} \times \sqrt{1-\frac{1}{5}}\right)$

$=\sin ^{-1}\left(\frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}} \times \frac{2}{\sqrt{5}}\right)$

$=\sin ^{-1}\left(\frac{1}{5}+\frac{4}{5}\right)$

$=\sin ^{-1} \frac{5}{5}$

$=\sin ^{-1} 1$

$=\frac{\pi}{2}$

$=\mathrm{RHS}$

Therefore, LHS = RHS

Hence proved.