Question

Prove the following results

(i) $\tan \left({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right)=\frac{17}{6}$
(ii) $\cos \left({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2}\right)=\frac{6}{5\sqrt{13}}$

(iii) $\tan \left({\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}\right)=\frac{63}{16}$

(iv) $\sin \left({\cos }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}\right)=\frac{63}{65}$

Answer 1

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

(ii)
$$\begin{gathered} \mathrm{LHS}=\cos \left({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2}\right) \\ =\cos \left({\sin }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{2}{3}\right) \\ =\cos \left[{\cos }^{-1}\sqrt{1-{\left(\frac{3}{5}\right)}^2}+{\cos }^{-1}\frac{1}{\sqrt{1+{\left({\displaystyle \frac{2}{3}}\right)}^2}}\right] \\ =\cos \left({\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{3}{\sqrt{13}}\right) \\ =\cos \left\{{\cos }^{-1}\left[\frac{4}{5}\times \frac{3}{\sqrt{13}}-\sqrt{1-{\left(\frac{4}{5}\right)}^2}\sqrt{1-{\left(\frac{3}{\sqrt{13}}\right)}^2}\right]\right\} \\ =\cos \left\{{\cos }^{-1}\left[\frac{12}{5\sqrt{13}}-\frac{6}{5\sqrt{13}}\right]\right\} \\ =\cos \left\{{\cos }^{-1}\left[\frac{6}{5\sqrt{13}}\right]\right\} \\ =\frac{6}{5\sqrt{13}}=\mathrm{RHS} \end{gathered}$$

(iii)
The question is wrong as we can't have arc sin greater than 1

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