√3+i=(a+ib)(c+id),thentan−1ba+tan−1dc has the value

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$\sqrt{3} + i = (a + i b) (c + i d), t h e n {tan}^{- 1} \frac{b}{a} + {tan}^{- 1} \frac{d}{c}$

has the value

Options:

A . 2n

π

+

\frac{π}{3}

, n

ϵ

I

B . n

π

+

\frac{π}{6}

, n

ϵ

I

C . n

π

-

\frac{π}{3}

, n

ϵ

I

D . 2n

π

-

\frac{π}{6}

, n

ϵ

I

Answer: Option B
:
B

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

$∴ a c - b d$ = $\sqrt{3}$ and ad+bc=1

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

= ${tan}^{- 1} (\frac{\frac{a}{b} + \frac{d}{c}}{1 - \frac{b}{a} . \frac{d}{c}})$ = ${tan}^{- 1} (\frac{b c + a d}{a c - b d})$ = ${tan}^{- 1} (\frac{1}{\sqrt{3}})$

=n $π$ + $\frac{π}{6}$ , n $ϵ$ I

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