Prove the following results:

Solution:

(i) LHS $=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}$

$=\tan ^{-1}\left(\frac{\frac{1}{\tau}+\frac{1}{13}}{1-\frac{1}{7} \times \frac{1}{13}}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{20}{91}}{\frac{90}{91}}\right)$

$=\tan ^{-1} \frac{2}{9}=$ RHS

(ii) LHS $=\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$

$=\tan ^{-1} \frac{\frac{12}{13}}{\sqrt{1-\frac{144}{165}}}+\tan ^{-1} \frac{\sqrt{1-\frac{16}{25}}}{\frac{4}{5}}+\tan ^{-1} \frac{63}{16} \quad\left[\because \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right.$ and $\left.\cos ^{-1} x=\tan ^{-1} \frac{\sqrt{1-x^{2}}}{x}\right]$

$=\tan ^{-1} \frac{\frac{12}{13}}{\frac{5}{13}}+\tan ^{-1} \frac{\frac{3}{5}}{\frac{4}{5}}+\tan ^{-1} \frac{63}{16}$            $\left[\because \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1} \frac{12}{5}+\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{63}{16}$

$=\pi+\tan ^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}\right)+\tan ^{-1} \frac{63}{16}$

$=\pi+\tan ^{-1} \frac{-63}{16}+\tan ^{-1} \frac{63}{16}$

$=\pi-\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{63}{16}$

$=\pi=\mathrm{RHS}$

(iii)

LHS $=\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$

$=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right)$             $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{17}{30}}{\frac{34}{30}}\right)$

$=\tan ^{-1} \frac{1}{2}$

$=\sin ^{-1} \frac{\frac{1}{2}}{\sqrt{1+\left(\frac{1}{2}\right)^{2}}}$

$=\sin ^{-1} \frac{1}{\sqrt{5}}=$ RHS