If sqrt 3 + i = (a + ib)(c + id), then {tan ^ | vedclass

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(b)We have $\sqrt 3 + i = (a + ib)(c + id)$
$	herefore ac - bd = \sqrt 3 $and $ad + bc = 1$
Now $tan^{-1}$ $\left( {\frac{b}{a}} ight) + {	an ^{

Vedclass · Accepted Answer

$n\pi + \frac{\pi }{6},n \in I$

Vedclass · Answer

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

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

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