Question

(i) ${\sin }^{-1}\frac{63}{65}={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}$
(ii) ${\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{63}{16}$
(iii) $\frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}$

Answer 1

(i)
$$\begin{gathered} \mathrm{RHS} \\ {\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5} \\ ={\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{4}{5}\left[\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right] \\ ={\sin }^{-1}\left\{\frac{5}{13}\sqrt{1-{\left(\frac{4}{5}\right)}^2}+\frac{4}{5}\sqrt{1-{\left(\frac{5}{13}\right)}^2}\right\} \\ ={\sin }^{-1}\left\{\frac{5}{13}\times \frac{3}{5}+\frac{4}{5}\times \frac{12}{13}\right\} \\ ={\sin }^{-1}\left\{\frac{15}{65}+\frac{48}{65}\right\} \\ ={\sin }^{-1}\frac{63}{65}=\mathrm{LHS} \end{gathered}$$


(ii)

$$\begin{gathered} \mathrm{LHS}={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5} \\ ={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5} \\ ={\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\left[\because {\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}\right] \\ ={\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{4}{5} \\ ={\sin }^{-1}\left[\frac{5}{13}\sqrt{1-{\left(\frac{4}{5}\right)}^2}+\frac{4}{5}\sqrt{1-{\left(\frac{5}{13}\right)}^2}\right]\left[\because {\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)\right] \\ ={\sin }^{-1}\left(\frac{5}{13}\times \frac{3}{5}+\frac{4}{5}\times \frac{12}{13}\right) \\ ={\sin }^{-1}\left(\frac{3}{13}+\frac{48}{65}\right) \\ ={\sin }^{-1}\left(\frac{63}{65}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{63}{65}}}{\sqrt{1-{{\displaystyle \frac{63}{65}}}^2}}\right)\left[\because {\sin }^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{63}{65}}}{{\displaystyle \frac{16}{65}}}\right) \\ ={\tan }^{-1}\left(\frac{63}{16}\right)=\mathrm{RHS} \end{gathered}$$

(iii)

$$\begin{gathered} \frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3} \\ \mathrm{LHS}=\frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3} \\ =\frac{9}{4}\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\frac{1}{3}\right) \\ =\frac{9}{4}\left({\cos }^{-1}\frac{1}{3}\right) \\ =\frac{9}{4}\left({\sin }^{-1}\sqrt{1-\frac{1}{9}}\right) \\ =\frac{9}{4}\left({\sin }^{-1}\frac{2\sqrt{2}}{3}\right)=\mathrm{RHS} \end{gathered}$$