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) $\sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\frac{63}{65}$
(i)
LHS $=\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\tan \left(\tan ^{-1} \frac{\sqrt{1-\left(\frac{4}{5}\right)^{2}}}{\frac{4}{5}}+\tan ^{-1} \frac{2}{3}\right) \quad\left[\because \cos ^{-1} x=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\right]$
$=\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$
$=\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)\right] \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$=\tan \left[\tan ^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)\right]$
$=\tan \left[\tan ^{-1} \frac{17}{6}\right]$
$=\frac{17}{6}=$ RHS
(ii)
LHS $=\cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$
$=\cos \left(\sin ^{-1} \frac{3}{5}+\tan ^{-1} \frac{2}{3}\right)$
$=\cos \left[\cos ^{-1} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\cos ^{-1} \frac{1}{\sqrt{1+\left(\frac{2}{3}\right)^{2}}}\right]$
$=\cos \left(\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{3}{\sqrt{13}}\right)$
$=\cos \left\{\cos ^{-1}\left[\frac{4}{5} \times \frac{3}{\sqrt{13}}-\sqrt{1-\left(\frac{4}{5}\right)^{2}} \sqrt{1-\left(\frac{3}{\sqrt{13}}\right)^{2}}\right]\right\}$
$=\cos \left\{\cos ^{-1}\left[\frac{12}{5 \sqrt{13}}-\frac{6}{5 \sqrt{13}}\right]\right\}$
$=\cos \left\{\cos ^{-1}\left[\frac{6}{5 \sqrt{13}}\right]\right\}$
$=\frac{6}{5 \sqrt{13}}=\mathrm{RHS}$
(iii)
The question is wrong as we can't have arc sin greater than 1
LHS $=\sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)$
$=\sin \left[\sin ^{-1} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\sin ^{-1} \frac{5}{13}\right]$
$=\sin \left[\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}\right]$
$=\sin \left\{\sin ^{-1}\left[\frac{4}{5} \times \sqrt{1-\left(\frac{5}{13}\right)^{2}}+\frac{5}{13} \times \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right]\right\}$
$=\sin \left[\sin ^{-1}\left(\frac{48}{65}+\frac{15}{65}\right)\right]$
$=\sin \left(\sin ^{-1} \frac{63}{65}\right)$
$=\frac{63}{65}=$ RHS