Prove that:

To Prove: $\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13}=\sin ^{-1} \frac{56}{65}$

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:

$\mathrm{LHS}=\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13} \ldots$ (1)

Let $\cos \theta=\frac{3}{5}$

Therefore $\theta=\cos ^{-1} \frac{3}{5} \ldots$ (2)

From the figure, $\sin \theta=\frac{4}{5}$

$\Rightarrow \theta=\sin ^{-1} \frac{4}{5} \ldots$ (3)

From $(2)$ and $(3)$,

$\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{4}{5}$

Substituting in (1), we get

$\mathrm{LHS}=\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{12}{13}$

$=\sin ^{-1}\left(\frac{4}{5} \times \sqrt{1-\left(\frac{12}{13}\right)^{2}}+\frac{12}{13} \times \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right)$

$=\sin ^{-1}\left(\frac{4}{5} \times \sqrt{1-\frac{144}{169}}+\frac{12}{13} \times \sqrt{\left.1-\frac{16}{25}\right)}\right.$

$=\sin ^{-1}\left(\frac{4}{5} \times \sqrt{\frac{25}{169}}+\frac{12}{13} \times \sqrt{\frac{9}{25}}\right)$

$=\sin ^{-1}\left(\frac{4}{5} \times \frac{5}{13}+\frac{12}{13} \times \frac{3}{5}\right)$

$=\sin ^{-1}\left(\frac{20}{65}+\frac{36}{65}\right)$

$=\sin ^{-1} \frac{56}{65}$

$=\mathrm{RHS}$

Therefore, LHS = RHS

Hence proved.