15.

Prove that   ${\sin }^{- 1} \left( \frac{8}{17} \right) + {\sin }^{- 1} \left( \frac{3}{5} \right) = {\cos }^{- 1} \left( \frac{36}{85} \right).$

$$\begin{gathered} \mathrm{Let} {\sin }^{- 1} \frac{8}{17} = \mathrm{x}. \\ \mathrm{Then}, \sin \mathrm{x} =\frac{8}{17}; \cos \mathrm{x} = \sqrt{ 1 - {\mathrm{x}}^2} \\ \Rightarrow \cos \mathrm{x} = \sqrt{ 1 - {\left( \frac{8}{17} \right)}^2} \\ \Rightarrow \cos \mathrm{x} = \sqrt{\frac{225}{289}} \\ \Rightarrow \cos \mathrm{x} = \frac{15}{17} \end{gathered}$$

$$\begin{gathered} \therefore \tan \mathrm{x} = \frac{\sin \mathrm{x}}{\cos \mathrm{x}} \\ \Rightarrow \tan \mathrm{x} = \frac{{\displaystyle \frac{8}{17}}}{{\displaystyle \frac{15}{17}}} \\ \Rightarrow \tan \mathrm{x} = \frac{8}{15} \\ \Rightarrow = \mathrm{x} = {\tan }^{- 1} \left( \frac{8}{15} \right) ...........( \mathrm{i} ) \\ \mathrm{Let} {\sin }^{- 1} \frac{3}{5} = \mathrm{y} ...........( \mathrm{ii} ) \\ \mathrm{Then}, \sin \mathrm{y} = \frac{3}{5}; \cos \mathrm{y} = \sqrt{ 1 - {\mathrm{y}}^2} \end{gathered}$$

$$\begin{gathered} \Rightarrow \cos \mathrm{y} = \sqrt{ 1 -{\left(\frac{3}{5}\right)}^2 } \\ \Rightarrow \cos \mathrm{y} = \sqrt{ \frac{16}{25} } \\ \Rightarrow \cos \mathrm{y} = \frac{4}{5} \\ \therefore \tan \mathrm{y} = \frac{\sin \mathrm{y}}{\cos \mathrm{y}} \\ \Rightarrow \tan \mathrm{y} = \frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}} \\ \Rightarrow \tan \mathrm{y} = \frac{3}{4} \\ \Rightarrow \mathrm{y} = {\tan }^{- 1} \left( \frac{3}{4} \right) ...............( \mathrm{iii} ) \end{gathered}$$

From equation ( ii )  and  ( iii ),  we have,

$$\begin{gathered} {\sin }^{- 1} \left( \frac{3}{5} \right) = {\tan }^{- 1} \left( \frac{3}{4} \right) \\ \mathrm{Now} \mathrm{consider} {\sin }^{- 1} \left( \frac{8}{17} \right) + {\sin }^{- 1} \left( \frac{3}{5} \right): \end{gathered}$$

From equation  ( i )  and  ( iii ), we have,

$$\begin{gathered} {\sin }^{- 1 }\left( \frac{8}{17} \right) + {\sin }^{- 1 }\left( \frac{3}{5} \right) = {\tan }^{- 1} \left( \frac{8}{15} \right) + {\tan }^{- 1} \left( \frac{3}{4} \right) \\ = {\tan }^{- 1} \left( \frac{{\displaystyle \frac{8}{15}} + {\displaystyle \frac{3}{4}}}{1 - {\displaystyle \frac{8}{15}} \times {\displaystyle \frac{3}{4}}} \right) .......\left[ \because {\tan }^{- 1} \mathrm{x} + {\tan }^{- 1} \mathrm{y} = {\tan }^{- 1} \frac{\mathrm{x} + \mathrm{y}}{1 - \mathrm{xy}} \right] \\ = {\tan }^{- 1} \left( \frac{32 + 45}{60 - 24} \right) \\ {\sin }^{- 1} \left( \frac{8}{17} \right) + {\sin }^{- 1} \left( \frac{3}{5} \right) = {\tan }^{- 1} \left( \frac{77}{36} \right) ........( \mathrm{iv} ) \end{gathered}$$

Now, we have:

$$\begin{gathered} \mathrm{Let} {\tan }^{- 1} \left( \frac{77}{36} \right) = \mathrm{z}. \\ \mathrm{Then} \tan \mathrm{z} = \frac{77}{36} \\ \Rightarrow \sec \mathrm{z} = \sqrt{ 1 + {\left(\frac{77}{36}\right)}^2} .......\left[ \because \sec \mathrm{\theta } = \sqrt{ 1 + {\tan }^2 \mathrm{\theta }} \right] \\ \Rightarrow \sec \mathrm{z} = \sqrt{ \frac{1296 + 5929}{1296}} \\ \Rightarrow \sec \mathrm{z} = \sqrt{ \frac{7225}{1296}} \\ \Rightarrow \sec \mathrm{z} = \frac{85}{36} \end{gathered}$$

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \cos \mathrm{z} = \frac{1}{\sec \mathrm{z}} \\ \mathrm{Thus}, \sec \mathrm{z} = \frac{85}{36}, \cos \mathrm{z} = \frac{36}{85} \\ \Rightarrow \mathrm{z} = {\cos }^{- 1} \left( \frac{36}{85} \right) \\ \Rightarrow {\tan }^{- 1} \left( \frac{77}{36} \right) = {\cos }^{- 1} \left( \frac{36}{85}\right) \\ \Rightarrow {\sin }^{- 1} \left( \frac{8}{17} \right) + {\sin }^{- 1} \left( \frac{3}{5} \right) = {\cos }^{- 1} \left( \frac{36}{85}\right) .......[ \because \mathrm{From} \mathrm{equation} ( \mathrm{iv} ) ] \end{gathered}$$

Hence proved.