Q. If $\tan{2x = \tan\frac{2}{x}}$, then the value of $x$ is
A) $\frac{n\pi \pm \sqrt{n^{2}\pi^{2} + 16}}{4}$
B) $\frac{n\pi}{4}$
C) $\frac{n\pi \pm \sqrt{n^{2}\pi^{2} - 16}}{4}$
D) None of these
Explanation
(a)
Since, $\tan{2x = \tan\frac{2}{x}}$
$\Rightarrow \ 2x = n\pi + \frac{2}{x} \Rightarrow 2x^{2} - n\pi x - 2 = 0$
$\Rightarrow \ x = \frac{n\pi \pm \sqrt{n^{2}\pi^{2} + 16}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.