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The general solution of \( \cot \theta+\tan \theta=2 \) is
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Verified Answer
The correct answer is:
\( \frac{n I}{2}+(-1)^{n \frac{n}{4}} \)
Given that, $\cot \theta+\tan \theta=2$
$\Rightarrow \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=2$
$\Rightarrow \frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=2$
$\Rightarrow \frac{1}{\sin \theta \cos \theta}=2$
$\Rightarrow 2 \sin \theta \cos \theta=1$
$\Rightarrow \sin 2 \theta=1=\sin \left(\frac{\Pi}{2}\right)$
As we know, if $\sin \theta=\sin \alpha$ then $\theta=n \Pi+(-1)^{n} \alpha$
Therefore, $2 \theta=n \Pi+(-1)^{n} \frac{\Pi}{2}$
$\Rightarrow \theta=\frac{n \Pi}{2}+(-1)^{n} \frac{\Pi}{4}$
$\Rightarrow \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=2$
$\Rightarrow \frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=2$
$\Rightarrow \frac{1}{\sin \theta \cos \theta}=2$
$\Rightarrow 2 \sin \theta \cos \theta=1$
$\Rightarrow \sin 2 \theta=1=\sin \left(\frac{\Pi}{2}\right)$
As we know, if $\sin \theta=\sin \alpha$ then $\theta=n \Pi+(-1)^{n} \alpha$
Therefore, $2 \theta=n \Pi+(-1)^{n} \frac{\Pi}{2}$
$\Rightarrow \theta=\frac{n \Pi}{2}+(-1)^{n} \frac{\Pi}{4}$
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