Question

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

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

Answer 1

The Correct Option is B

Given: $\cot \theta +\tan \theta =2$
$\Rightarrow \frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=2,$
use basic trigonometric identities
$\Rightarrow \frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=2$$\Rightarrow \frac{1}{\sin \theta \cos \theta }=2,$
Pythagorean identity$\Rightarrow 2\sin \theta \cos \theta =1$$\Rightarrow \sin 2\theta =1,$
double angle formula$\Rightarrow \sin 2\theta =\sin \frac{\pi }{2}$
Hence general solution is given by,
$2\theta =n\pi +(-1{)}^n\frac{\pi }{2},$ where $n\in Z$$\therefore \theta =\frac{\mathrm{n}\pi }{2}+(-1{)}^{\mathrm{n}}\frac{\pi }{4}$
Note that: If $\sin \theta =\sin \alpha ,$
then general solution is given by
$\theta =n\pi +(-1{)}^n\alpha$
Hence, the correct option is (B).