Let  a  be in I quadrant  such that

$$\begin{gathered} {\cos }^{-1 }\left(\frac{12}{13}\right) = \mathrm{a} \\ \mathrm{So} \cos \mathrm{a} = \frac{12}{13} \\ \Rightarrow \sin \mathrm{a} = \sqrt{ 1 - {\left(\frac{12}{13}\right)}^2} \\ = \sqrt{ 1 - \frac{144}{169}} \\ = \sqrt{ \frac{169 - 144}{169}} \\ = \sqrt{\frac{25}{169}} = \frac{5}{13} \\ \mathrm{And} \tan \mathrm{a} = \frac{5}{12} \\ \mathrm{So}, \mathrm{a} = {\tan }^{-1} \left( \frac{5}{12} \right) .........\left(\mathrm{i}\right) \\ \mathrm{Again} \mathrm{b}\in \mathrm{I} \mathrm{quadrant} \mathrm{such} \mathrm{that} {\sin }^{-1} \left( \frac{3}{5} \right) = \mathrm{b} \\ \mathrm{So}, \sin \mathrm{b} = \frac{3}{5} \end{gathered}$$

$$\begin{gathered} \Rightarrow \cos \mathrm{b} = \sqrt{ 1 - {\left( \frac{3}{5} \right)}^2} \\ = \sqrt{ 1 - \frac{9}{25}} \\ = \sqrt{\frac{16}{25}} = \frac{4}{5} \\ \mathrm{And} \tan \mathrm{b} = \frac{3}{4} \\ \mathrm{so}, \mathrm{b} = {\tan }^{-1} \left( \frac{3}{4} \right) ...............\left(\mathrm{ii}\right) \\ \mathrm{Now}, \mathrm{let} {\sin }^{-1} \left( \frac{56}{65} \right) = \mathrm{c} \mathrm{where} \mathrm{c} \mathrm{is} \mathrm{in} \mathrm{I} \mathrm{quadrant} \\ \mathrm{So}, \sin \mathrm{c} = \frac{56}{65} \end{gathered}$$

$$\begin{gathered} \Rightarrow \cos \mathrm{c} = \sqrt{ 1 - {\left( \frac{56}{65} \right)}^2} \\ = \sqrt{ 1 - \frac{3136}{4225}} \\ = \sqrt{\frac{4225 - 3136}{4225}} \\ = \sqrt{\frac{1089}{4225}} = \frac{33}{65} \\ \mathrm{And}, \tan \mathrm{c} = \frac{56}{33} \\ \mathrm{So}, \mathrm{c} = {\tan }^{-1} \left( \frac{56}{33} \right) \\ \Rightarrow {\sin }^{-1} \left( \frac{56}{65} \right) = {\tan }^{-1} \left( \frac{56}{33} \right) ...........\left(\mathrm{iii}\right) \\ \mathrm{Now}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{prove} {\cos }^{-1} \left(\frac{12}{13}\right) + {\sin }^{-1} \left(\frac{3}{5}\right) = {\sin }^{-1} \left( \frac{56}{65} \right) \end{gathered}$$

consider  a + b

$$\begin{gathered} = {\cos }^{-1} \left(\frac{12}{13}\right) + {\sin }^{-1} \left( \frac{3}{5} \right) \\ = {\tan }^{-1} \left(\frac{5}{12}\right) + {\tan }^{-1} \left( \frac{3}{4} \right) \left[\begin{array}{c} {\cos }^{-1 } \left(\frac{12}{13}\right) = {\tan }^{-1} \left(\frac{5}{12}\right) \mathrm{and} \\ {\sin }^{-1} \left( \frac{3}{5} \right) = {\tan }^{-1} \left( \frac{3}{4} \right)\end{array}\right] \\ = {\tan }^{-1} \left[ \frac{{\displaystyle \frac{5}{12}} + {\displaystyle \frac{3}{4}}}{1 - \left( {\displaystyle \frac{5}{12}} \mathrm{x} {\displaystyle \frac{3}{4}} \right)}\right] \left[ \mathrm{using}, {\tan }^{-1} \mathrm{x} +{\tan }^{-1} \mathrm{y} = {\tan }^{-1 }\left(\frac{ \mathrm{x} + \mathrm{y}}{ 1 - \mathrm{xy}} \right) \right] \\ = \tan { }^{-1} \left( \frac{20 + 36}{48 - 15} \right) \\ = \tan { }^{-1} \left( \frac{56}{33} \right) \\ = \mathrm{c} = {\sin }^{-1} \left( \frac{56}{65} \right) \left[ \mathrm{using}, \mathrm{eq}, \left(\mathrm{iii}\right) \right] \end{gathered}$$

Hence proved.