If 1 + cot theta = text{cosec} theta,then the | vedclass

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(C) Given equation: $1 + \cot 	heta = 	ext{cosec} 	heta$Rewrite in terms of $\sin 	heta$ and $\cos 	heta$: $1 + \frac{\cos 	heta}{\sin 	heta} =

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$2n\pi + \frac{\pi}{2}$

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(C) Given equation: $1 + \cot 	heta = 	ext{cosec} 	heta$Rewrite in terms of $\sin 	heta$ and $\cos 	heta$: $1 + \frac{\cos 	heta}{\sin 	heta} = \frac{1}{\sin 	heta}$Multiply by $\sin 	heta$ (where $\sin 	heta 
eq 0$): $\sin 	heta + \cos 	heta = 1$Divide by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta = \frac{1}{\sqrt{2}}$$\cos \frac{\pi}{4} \sin 	heta + \sin \frac{\pi}{4} \cos 	heta = \frac{1}{\sqrt{2}}$$\sin(	heta + \frac{\pi}{4}) = \sin \frac{\pi}{4}$General solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$:Case $1$: If $n$ is even,let $n = 2k$: $	heta + \frac{\pi}{4} = 2k\pi + \frac{\pi}{4} \Rightarrow 	heta = 2k\pi$Case $2$: If $n$ is odd,let $n = 2k+1$: $	heta + \frac{\pi}{4} = (2k+1)\pi - \frac{\pi}{4}$ $\Rightarrow 	heta = 2k\pi + \pi - \frac{\pi}{2} = 2k\pi + \frac{\pi}{2}$Check domain: If $	heta = 2k\pi$,then $\cot 	heta$ and $	ext{cosec} 	heta$ are undefined. Thus,$	heta = 2n\pi$ is rejected.Therefore,the general solution is $	heta = 2n\pi + \frac{\pi}{2}$.

If $1 + \cot \theta = \text{cosec} \theta$,then the general value of $\theta$ is

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