Question

If $\sqrt{3}\tan 2\mathrm{\theta }+\sqrt{3}\tan 3\mathrm{\theta }+\tan 2\mathrm{\theta }\tan 3\mathrm{\theta }=1$, then the general value of $\theta$ is

• A. $n\pi +\frac{\pi }{5}$
• B. $\left(n+\frac{1}{6}\right)\frac{\pi }{5}$
• C. $\left(2n\pm \frac{1}{6}\right)\frac{\pi }{5}$
• D. $\left(n+\frac{1}{3}\right)\frac{\pi }{5}$

Answer 1

The correct option is B

$\left(n+\frac{1}{6}\right)\frac{\pi }{5}$

Explanation for the correct option:

Step 1: Rearranging as a form of the formula,

Given $\sqrt{3}\tan 2\mathrm{\theta }+\sqrt{3}\tan 3\mathrm{\theta }+\tan 2\mathrm{\theta }\tan 3\mathrm{\theta }=1$

$\begin{array}{l}\Rightarrow \sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1\\ \Rightarrow \sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta =1-\tan 2\theta \tan 3\theta \\ \Rightarrow \sqrt{3}(\tan 2\theta +\tan 3\theta )=1-\tan 2\theta \tan 3\theta \\ \Rightarrow \frac{(\tan 2\theta +\tan 3\theta )}{1-\tan 2\theta \tan 3\theta }=\frac{1}{\sqrt{3}}.......\left(i\right)\end{array}$

Step 2: Apply the formula $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ and $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$ ,

Here, Let $A=2\theta$ and $B=3\theta$, then equation becomes as follows

$\tan (2\theta +3\theta )=\tan \frac{\pi }{6}$,

As per, the trigonometric rule if $\tan x=\tan y$ then the general solution is $x=n\mathrm{\pi }+y$, then the above equation becomes

$\begin{array}{l}\Rightarrow 2\theta +3\theta =n\pi +\frac{\pi }{6}\\ \Rightarrow 5\theta =n\pi +\frac{\pi }{6}\\ \Rightarrow \theta =\frac{1}{5}\left(n\pi +\frac{\pi }{6}\right)\\ \Rightarrow \theta =\frac{\pi }{5}\left(n\pi +\frac{\pi }{6}\right)\end{array}$

Therefore, the general solution of $\theta$ is $\frac{\pi }{5}\left(n+\frac{1}{6}\right)$.

Hence, the correct option is (B).