Explanation

(a)
Let $\cot^{- 1}{\frac{3}{4} = \theta \Rightarrow \cot{\theta = \frac{3}{4}}}$
And $\sin{\theta = \frac{1}{\sqrt{1 + \cot^{2}\theta}} = \frac{1}{\sqrt{1 + \left( \frac{9}{16} \right)}} = \frac{4}{5}}$
$\therefore\cot^{- 1}{\frac{3}{4} + \sin^{- 1}{\frac{5}{13} = \sin^{- 1}\frac{4}{5}} + \sin^{- 1}\frac{5}{13}}$
$= \sin^{- 1}\left\lbrack \frac{4}{5}\sqrt{1 - \frac{25}{269}} + \frac{5}{13}\sqrt{1 - \frac{16}{25}} \right\rbrack$
${= sin}^{- 1}\left\lbrack \frac{4}{5} \bullet \frac{12}{13} + \frac{5}{13} \bullet \frac{3}{5} \right\rbrack$
$= \sin^{- 1}{\left\lbrack \frac{48 + 15}{65} \right\rbrack = \sin^{- 1}\frac{63}{65}}$