To Prove:

$$\cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{56}{65}\right)$$

Proof:

Let $\cos^{-1}\left(\frac{12}{13}\right) = \theta_1$ and $\sin^{-1}\left(\frac{3}{5}\right) = \theta_2$.

Step 1: Find $\sin(\theta_1)$ and $\cos(\theta_2)$.

Since $\cos^{-1}\left(\frac{12}{13}\right) = \theta_1$, we have $\cos(\theta_1) = \frac{12}{13}$.

Therefore,

\sin(\theta_1) = \sqrt{1 - \cos^2\theta_1} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \textbf{\frac{5}{13}}

Since $\sin^{-1}\left(\frac{3}{5}\right) = \theta_2$, we have $\sin(\theta_2) = \frac{3}{5}$.

Therefore,

\cos(\theta_2) = \sqrt{1 - \sin^2\theta_2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \textbf{\frac{4}{5}}

Step 2: Apply the addition formula for sine.

$$\sin(\theta_1 + \theta_2) = \sin(\theta_1)\cos(\theta_2) + \cos(\theta_1)\sin(\theta_2)$$ $$= \left(\frac{5}{13}\right) \times \left(\frac{4}{5}\right) + \left(\frac{12}{13}\right) \times \left(\frac{3}{5}\right)$$ $$= \frac{20}{65} + \frac{36}{65}$$ = \textbf{\frac{56}{65}}

Step 3: Verify the range.

Since $\frac{12}{13} > 0$, $\theta_1 = \cos^{-1}\left(\frac{12}{13}\right)$ lies in $\left(0, \frac{\pi}{2}\right)$.

Since $\frac{3}{5} > 0$, $\theta_2 = \sin^{-1}\left(\frac{3}{5}\right)$ lies in $\left(0, \frac{\pi}{2}\right)$.

Thus $\theta_1 + \theta_2$ lies in $(0, \pi)$, which is within the principal range of $\sin^{-1}$.

Therefore, $\theta_1 + \theta_2 = \sin^{-1}\left(\frac{56}{65}\right)$

Hence proved:

$$\cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{56}{65}\right)$$