The general solution of + =2 is - Vidyadip

MCQ

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

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

✓

Answer

Correct option: A.

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

(A) $\cot \theta+\tan \theta=2$
$\begin{array}{l}\therefore \quad \frac{1}{\tan \theta}+\tan \theta=2 \Rightarrow 1+\tan ^2 \theta=2 \tan \theta \\ \therefore \quad \frac{2 \tan \theta}{1+\tan ^2 \theta}=1 \Rightarrow \sin 2 \theta=1\end{array}$
$\begin{array}{l}\Rightarrow 2 \theta= n \pi+(-1)^{ n } \frac{\pi}{2} \\ \Rightarrow \theta=\frac{ n \pi}{2}+(-1)^{ n } \frac{\pi}{4}\end{array}$

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