Question

Prove the following results:
(i) ${\cos }^{-1}\left(\frac{5}{13}\right)={\tan }^{-1}\left(\frac{12}{5}\right)$
(ii) ${\sin }^{-1}\left(-\frac{4}{5}\right)={\tan }^{-1}\left(-\frac{4}{3}\right)={\cos }^{-1}\left(-\frac{3}{5}\right)-\mathrm{\pi }$
(iii) $\tan \left({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right)=\frac{17}{6}$
(iv) $2{\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{24}{7}$
(v) ${\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{63}{16}$

Answer 1

(i)
$$\begin{gathered} \mathrm{LHS}={\cos }^{-1}\frac{5}{13}={\tan }^{-1}\frac{\sqrt{1-{\left({\displaystyle \frac{5}{13}}\right)}^2}}{{\displaystyle \frac{5}{13}}}\left[\because {\cos }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right] \\ ={\tan }^{-1}\frac{{\displaystyle \frac{12}{13}}}{{\displaystyle \frac{5}{13}}} \\ ={\tan }^{-1}\frac{12}{5}=\mathrm{RHS} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{LHS}={\sin }^{-1}-\left(\frac{4}{5}\right)={\tan }^{-1}\frac{-{\displaystyle \frac{4}{5}}}{{\displaystyle \sqrt{1-{\left(\frac{4}{5}\right)}^2}}}\left[\because {\sin }^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right] \\ ={\tan }^{-1}-\frac{{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}} \\ ={\tan }^{-1}-\frac{4}{3}=\mathrm{RHS} \\ \mathrm{Again}, \\ {\sin }^{-1}-\left(\frac{4}{5}\right)=-{\sin }^{-1}\left(\frac{4}{5}\right) \\ =-{\cos }^{-1}\sqrt{1-{\left(\frac{4}{5}\right)}^2}\left[\because {\sin }^{-1}x={\cos }^{-1}\left(\sqrt{1-x^2}\right)\right] \\ =-{\cos }^{-1}\frac{3}{5} \\ =-\left[\mathrm{\pi }-{\cos }^{-1}\left(-\frac{3}{5}\right)\right] \\ ={\cos }^{-1}\left(-\frac{3}{5}\right)-\mathrm{\pi } \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{LHS}=\tan \left({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right)=\tan \left({\tan }^{-1}\frac{\sqrt{1-{\left({\displaystyle \frac{4}{5}}\right)}^2}}{{\displaystyle \frac{4}{5}}}+{\tan }^{-1}\frac{2}{3}\right)\left[\because {\cos }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right] \\ =\tan \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right) \\ =\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{2}{3}}}{1-{\displaystyle \frac{3}{4}\times \frac{2}{3}}}\right)\right]\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ =\tan \left[{\tan }^{-1}\left(\frac{{\displaystyle \frac{17}{12}}}{{\displaystyle \frac{6}{12}}}\right)\right] \\ =\tan \left[{\tan }^{-1}\frac{17}{6}\right] \\ =\frac{17}{6}=\mathrm{RHS} \end{gathered}$$


(iv)

$$\begin{gathered} \mathrm{LHS}=2{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\left(2\times \frac{3}{5}\sqrt{1-\frac{9}{25}}\right)\left[\because 2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)\right] \\ ={\sin }^{-1}\left(\frac{6}{5}\times \frac{4}{5}\right) \\ ={\sin }^{-1}\left(\frac{24}{25}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{24}{25}}}{\sqrt{1-{\left(\frac{24}{25}\right)}^2}}\right)\left[\because {\sin }^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{24}{25}}}{{\displaystyle \frac{7}{25}}}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle 24}}{7}\right)=\mathrm{RHS} \end{gathered}$$

(v)

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