• question_answer If $\sqrt{3}+i=(a+ib)(c+id)$, then ${{\tan }^{-1}}\left( \frac{b}{a} \right)+$ ${{\tan }^{-1}}\left( \frac{d}{c} \right)$ has the value A) $\frac{\pi }{3}+2n\pi ,n\in I$ B) $n\pi +\frac{\pi }{6},n\in I$ C) $n\pi -\frac{\pi }{3},n\in I$ D) $2n\pi -\frac{\pi }{3},n\in I$

We have $\sqrt{3}+i=(a+ib)(c+id)$       $\therefore ac-bd=\sqrt{3}$and $ad+bc=1$ Now  tan-1$\left( \frac{b}{a} \right)+{{\tan }^{-1}}\left( \frac{d}{c} \right)$ \begin{align} & ={{\tan }^{-1}}\left( \frac{\frac{b}{a}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}} \right)={{\tan }^{-1}}\left( \frac{bc+ad}{ac-bd} \right)={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right) \\ & \\ \end{align} $=n\pi +\frac{\pi }{6},n\in I$