Question

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.
• B. (n\pi \frac{\pi}{6},n~\epsilon~I\)
• C.
• D.

Answer 1

The correct option is C

We have $\sqrt{3}+i=(a+ib)(c+id)$
$\therefore ac-bd=\sqrt{3}andad+bc=1$
Now tan ${}^{-1}\left(\frac{b}{a}\right)+ta{n}^{-1}\left(\frac{d}{c}\right)$
= tan ${}^{-1}\left(\frac{\frac{b}{a}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}}\right)=ta{n}^{-1}\left(\frac{bc+ad}{ac-bd}\right)=ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$=n\pi +\frac{\pi }{6},nϵI$