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

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(B) Given $\sqrt{3} + i = (a + ib)(c + id)$.Expanding the right side,we get $\sqrt{3} + i = (ac - bd) + i(ad + bc)$.Comparing real and imaginary parts

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

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(B) Given $\sqrt{3} + i = (a + ib)(c + id)$.Expanding the right side,we get $\sqrt{3} + i = (ac - bd) + i(ad + bc)$.Comparing real and imaginary parts,we have $ac - bd = \sqrt{3}$ and $ad + bc = 1$.We need to find the value of $	heta = 	an^{-1}\left(\frac{b}{a}ight) + 	an^{-1}\left(\frac{d}{c}ight)$.Using the formula $	an^{-1}(x) + 	an^{-1}(y) = n\pi + 	an^{-1}\left(\frac{x+y}{1-xy}ight)$,we have:$	heta = n\pi + 	an^{-1}\left(\frac{\frac{b}{a} + \frac{d}{c}}{1 - \frac{b}{a} \cdot \frac{d}{c}}ight) = n\pi + 	an^{-1}\left(\frac{bc + ad}{ac - bd}ight)$.Substituting the values,$	heta = n\pi + 	an^{-1}\left(\frac{1}{\sqrt{3}}ight)$.Since $	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$,the expression equals $n\pi + \frac{\pi}{6}$ for $n \in I$.

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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