The general solution of cot theta + tan theta | vedclass

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Question

(B) Given that,$\cot 	heta + 	an 	heta = 2$ $\Rightarrow \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = 2$ $\Rightarrow \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given that,$\cot 	heta + 	an 	heta = 2$ $\Rightarrow \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = 2$ $\Rightarrow \frac{\cos^2 	heta + \sin^2 	heta}{\sin 	heta \cos 	heta} = 2$ $\Rightarrow \frac{1}{\sin 	heta \cos 	heta} = 2$ $\Rightarrow 2 \sin 	heta \cos 	heta = 1$ $\Rightarrow \sin 2 	heta = 1 = \sin \left(\frac{\pi}{2}ight)$ As we know,if $\sin x = \sin \alpha$,then $x = n \pi + (-1)^n \alpha$ Therefore,$2 	heta = n \pi + (-1)^n \frac{\pi}{2}$ $\Rightarrow 	heta = \frac{n \pi}{2} + (-1)^n \frac{\pi}{4}$

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

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