Question

$\sqrt{3}+i=(a+ib)(c+id),then{\tan }^{-1}\frac{b}{a}+{\tan }^{-1}\frac{d}{c}$

has the value

• A. 2n$\pi$+$\frac{\pi }{3}$, n$ϵ$I
• B. n$\pi$+$\frac{\pi }{6}$, n$ϵ$I
• C. n$\pi$-$\frac{\pi }{3}$, n$ϵ$I
• D. 2n$\pi$-$\frac{\pi }{6}$, n$ϵ$I

Answer 1

The correct option is B

n$\pi$+$\frac{\pi }{6}$, n$ϵ$I

$\sqrt{3}+i$=(a+ib)(c+id)

$\therefore ac-bd$=$\sqrt{3}$ and ad+bc=1

Now ${\tan }^{-1}\left(\frac{b}{a}\right)+{\tan }^{-1}\left(\frac{d}{c}\right)$

= ${\tan }^{-1}\left(\frac{\frac{a}{b}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}}\right)$= ${\tan }^{-1}\left(\frac{bc+ad}{ac-bd}\right)$= ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

=n$\pi$+$\frac{\pi }{6}$, n$ϵ$I