Question

If $\frac{\left(\tan \left(3\theta \right)-1\right)}{\left(\tan \left(3\theta \right)+1\right)}=\sqrt{3}$, then the general value of $\theta$ is

• A. $\frac{n\pi }{3}-\frac{\pi }{12}$
• B. $n\pi +\frac{7\pi }{12}$
• C. $\frac{n\pi }{3}+\frac{7\pi }{36}$
• D. $n\pi +\frac{\pi }{12}$

Answer 1

The correct option is C

$\frac{n\pi }{3}+\frac{7\pi }{36}$

Explanation for the correct option:

Step 1: Using componendo and dividendo rule

Given equation : $\frac{\left(\tan \left(3\theta \right)-1\right)}{\left(\tan \left(3\theta \right)+1\right)}=\sqrt{3}$

It can be expressed as $\mathit{a}\mathbf{:}\mathit{b}\mathbf{=}\mathit{c}\mathbf{:}\mathit{d}\mathbf{}\mathrm{and}\mathbf{}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\mathbf{:}\mathbf{(}\mathit{a}\mathbf{–}\mathit{b}\mathbf{)}\mathbf{=}\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathbf{)}\mathbf{:}\mathbf{(}\mathit{c}\mathbf{–}\mathit{d}\mathbf{)}$

$\begin{array}{rcl}\therefore \frac{\left[\tan \left(3\theta \right)-1+\tan \left(3\theta \right)+1\right]}{\left[\tan \left(3\theta \right)-1-\tan \left(3\theta \right)-1\right]}& =& \frac{\sqrt{3}+1}{\sqrt{3}-1}\\ & \Rightarrow & \frac{2\tan \left(3\theta \right)}{-2}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\ & \Rightarrow & \tan \left(3\theta \right)=\frac{\sqrt{3}+1}{1-\sqrt{3}}\end{array}$

Step 2: Applying the formula $\begin{array}{rcl}\mathbf{tan}\frac{\mathbf{\pi }}{\mathbf{3}}& \mathbf{=}& \sqrt{\mathbf{3}}\mathrm{and}\mathbf{}\mathbf{tan}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\mathbf{1}\end{array}$

$\begin{array}{rcl}\therefore \tan \left(3\theta \right)& =& \frac{\tan {\displaystyle \frac{\pi }{3}}+\tan {\displaystyle \frac{\pi }{4}}}{1-\tan \frac{\pi }{3}\tan \frac{\pi }{4}}\\ & =& \tan \left(\frac{\pi }{3}+\frac{\pi }{4}\right)\\ & =& \tan \frac{7\pi }{12}\end{array}$

$\begin{array}{rcl}\mathrm{Now},3\theta & =& n\pi +\frac{7\pi }{12}\\ \therefore \theta & =& \frac{n\pi }{3}+\frac{7\pi }{36}\end{array}$

Hence, the correct option is (C).