Question

If $\surd 3+i=(a+ib)(c+id)$, then ${\tan }^{-1}\left(\frac{b}{a}\right)+{\tan }^{-1}\left(\frac{d}{c}\right)$ has the value

• A. $\left(\frac{\pi }{3}\right)+2n\pi ,n\in I$
• B. $n\pi +\left(\frac{\pi }{6}\right),n\in I$
• C. $n\pi -\left(\frac{\pi }{3}\right),n\in I$
• D. $2n\pi -\left(\frac{\pi }{3}\right),n\in I$

Answer 1

The correct option is B

$n\pi +\left(\frac{\pi }{6}\right),n\in I$

Explanation for the correct option:

Step 1: Solve $\surd 3+i=(a+ib)(c+id)$

Given, $\surd 3+i=(a+ib)(c+id)$

$\Rightarrow$ $\surd 3+i=ac-bd+i(bc+ad)$

$\Rightarrow$ $ac–bd=\surd 3,bc+ad=i$

Step 2: Using the formula $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{A}\mathbf{}\mathbf{+}\mathbf{}\mathbf{B}}{\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{AB}}\mathbf{\right)}$

$\begin{array}{rcl}{\tan }^{-1}\left(\frac{b}{a}\right)+{\tan }^{-1}\left(\frac{d}{c}\right)& =& {\tan }^{-1}\left[\frac{\left(\frac{b}{a}\right)+\left(\frac{d}{c}\right)}{1-\left(\frac{b}{a}\right)\left(\frac{d}{c}\right)}\right]\\ & =& {\tan }^{-1}\left[\frac{{\displaystyle \frac{(bc+ad)}{ac}}}{{\displaystyle \frac{(ac–bd)}{ac}}}\right]\\ & =& {\tan }^{-1}\left(\frac{bc+ad}{ac–bd}\right)\\ & =& {\tan }^{-1}\left(\frac{1}{\surd 3}\right)\\ & =& \frac{\pi }{6}\end{array}$

$\mathbf{\therefore }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathbf{\pi }\mathbf{}\mathbf{+}\mathbf{}\left(\frac{\pi }{6}\right)\mathbf{,}\mathbf{}\mathbf{n}\mathbf{\in }\mathbf{I}$

Hence, Option $\text{'}B\text{'}$ is Correct.