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)

$\theta =\frac{n \pi} {2}+(-1)^n \frac {\pi}{4}$

Answer 2

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