If cot frac{x}{2} - operatorname{cosec} frac{ | vedclass

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Question

(A) Given the equation: $\cot \frac{x}{2} - \operatorname{cosec} \frac{x}{2} = \cot x$. Using the identities $\cot 	heta = \frac{\cos 	heta}{\sin

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$2 n \pi$

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(A) Given the equation: $\cot \frac{x}{2} - \operatorname{cosec} \frac{x}{2} = \cot x$. Using the identities $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$,we have: $\frac{\cos(x/2) - 1}{\sin(x/2)} = \frac{\cos x}{\sin x}$. Using $\cos(x/2) - 1 = -2 \sin^2(x/4)$ and $\sin(x/2) = 2 \sin(x/4) \cos(x/4)$,the left side becomes: $\frac{-2 \sin^2(x/4)}{2 \sin(x/4) \cos(x/4)} = -	an(x/4)$. The right side is $\cot x = \frac{1}{	an x}$. So,$-	an(x/4) = \frac{1}{	an x}$,which implies $	an x \cdot 	an(x/4) = -1$. This equation has no real solutions for $x$ because $	an x$ and $	an(x/4)$ cannot satisfy this product condition for all $n \in \mathbb{Z}$. However,checking the options provided,none satisfy the equation. Given the standard form of such problems,there is no solution.

If $\cot \frac{x}{2} - \operatorname{cosec} \frac{x}{2} = \cot x$,then the values of $x$ are

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