Question

General solution of the equation $\tan \theta \tan ({60}^{\circ }+\theta )\tan ({60}^{\circ }-\theta )=\sqrt{2}+1$ is

• A. $n\pi +\frac{\pi }{8},n\in Z$
• B. $\frac{n\pi }{3}+\frac{3\pi }{8},n\in Z$
• C. $n\pi +\frac{3\pi }{8},n\in Z$
• D. $\frac{n\pi }{3}+\frac{\pi }{8},n\in Z$

Answer 1

The correct option is D $\frac{n\pi }{3}+\frac{\pi }{8},n\in Z$

$\tan \theta \tan ({60}^{\circ }+\theta )\tan ({60}^{\circ }-\theta )=\sqrt{2}+1$
$\Rightarrow \tan 3\theta =\sqrt{2}+1=\tan \frac{3\pi }{8}$

So, the general solution of the above equation is
$3\theta =n\pi +\frac{3\pi }{8},n\in Z$
$\Rightarrow \theta =\frac{n\pi }{3}+\frac{\pi }{8},n\in Z$