Question

General solution of the equation $4cot2\theta +ta{n}^2\theta =co{t}^2\theta$ is (where, $n\in Z$ )

• A. $n\pi \pm \frac{\pi }{4}$
• B. $n\pi \pm \frac{\pi }{3}$
• C. $2n\pi \pm \frac{\pi }{3}$
• D. $2n\pi \pm \frac{\pi }{6}$

Answer 1

Answer: (A)

The given equation can be written as $\frac{2\left(1-{\tan }^2\theta \right)}{\tan \theta }+{\tan }^2\theta =\frac{1}{{\tan }^2\theta }$
$\Rightarrow \frac{2\left(1-{\tan }^2\theta \right)+{\tan }^3\theta }{\tan \theta }=\frac{1}{{\tan }^2\theta }$
$\Rightarrow 2\left(1-{\tan }^2\theta \right)\tan \theta +{\tan }^4\theta -1=0$
$\Rightarrow \left(1-{\tan }^2\theta \right)\left(2\tan \theta -1-{\tan }^2\theta \right)=0$
$\Rightarrow \left(1-{\tan }^2\theta \right)(1-\tan \theta {)}^2=0$
$\Rightarrow (1+\tan \theta )(1-\tan \theta {)}^3=0$
$\Rightarrow \tan \theta =\pm 1$
$\theta =n\pi \pm \frac{\pi }{4}$