Question

The general solution of equation $4\cot 2\theta ={\cot }^2\theta -{\tan }^2\theta$ is

• A. $n\pi \pm \frac{2\pi }{3},nϵz$
• B. $n\pi \pm \frac{\pi }{3},nϵz$
• C. $n\pi \pm \frac{\pi }{4},nϵz$
• D. None of these

Answer 1

Answer: (C)

$n\pi \pm \frac{\pi }{4},nϵz$

Given, $4\cot 2\theta ={\cot }^2\theta -{\tan }^2\theta$
$4(1-{\tan }^2\theta )\tan \theta =2(1-{\tan }^2\theta )$
$4\tan \theta -4{\tan }^3\theta =2-2{\tan }^4\theta$
${\tan }^4\theta -2{\tan }^3\theta +2\tan \theta -1=0$
$\tan \theta =1,1,1,-1$
$\tan \theta =\tan \frac{\pi }{4}$ or $\tan \frac{3\pi }{4}$
$\theta =n\pi \pm \frac{\pi }{4}$
$\theta =n\pi \pm \frac{3\pi }{4}$

so,
LCM $\Rightarrow$
$\theta =n\pi \pm \frac{\pi }{4}$