Question

The general solution of $\cot \theta +\tan \theta =2$ is

• A. $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{8}$
• B. $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4}$
• C. $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{6}$
• D. $\theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{8}$

Answer 1

Answer: (B)

We have, $\cot \theta +\tan \theta =2$
$\Rightarrow \frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=2$
$\Rightarrow \frac{{\cos }^2\theta +{\sin }^2\theta }{\sin \theta \cos \theta }=2$
$\Rightarrow 1=2\sin \theta \cos \theta$
$\Rightarrow \sin 2\theta =1$
$\Rightarrow \sin 2\theta =\sin \frac{\pi }{2}$
$\Rightarrow 2\theta =n\pi +(-1{)}^n\frac{\pi }{2}$
$\therefore \theta =\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4}$