If x neq 0,then frac{sin (pi+x) cos (frac{pi} | vedclass

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(C) Using the allied angle formulas: $\sin(\pi+x) = -\sin x$ $\cos(\frac{\pi}{2}+x) = -\sin x$ $	an(\frac{3\pi}{2}-x) = \cot x$ $\cot(2\pi-x) = -\cot

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Using the allied angle formulas: $\sin(\pi+x) = -\sin x$ $\cos(\frac{\pi}{2}+x) = -\sin x$ $	an(\frac{3\pi}{2}-x) = \cot x$ $\cot(2\pi-x) = -\cot x$ $\sin(2\pi-x) = -\sin x$ $\cos(2\pi+x) = \cos x$ $\operatorname{cosec}(-x) = -\operatorname{cosec} x$ $\sin(\frac{3\pi}{2}+x) = -\cos x$ Substituting these values into the expression: $= \frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x)}$ $= \frac{-\sin^2 x \cot^2 x}{\sin x \cos x \operatorname{cosec} x \cos x}$ $= \frac{-\sin^2 x \cot^2 x}{\sin x \cos^2 x \cdot \frac{1}{\sin x}}$ $= \frac{-\sin^2 x \cot^2 x}{\cos^2 x} = -	an^2 x \cdot \cot^2 x = -1$ Wait,let us re-evaluate the signs carefully: Numerator: $(-\sin x)(-\sin x)(\cot x)(-\cot x) = -\sin^2 x \cot^2 x$ Denominator: $(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x) = -\sin x \cos^2 x \operatorname{cosec} x = -\cos^2 x$ Result: $\frac{-\sin^2 x \cot^2 x}{-\cos^2 x} = \frac{\sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x}}{\cos^2 x} = \frac{\cos^2 x}{\cos^2 x} = 1$ Thus,the correct option is $C$.

If $x \neq 0$,then $\frac{\sin (\pi+x) \cos (\frac{\pi}{2}+x) \tan (\frac{3 \pi}{2}-x) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \operatorname{cosec}(-x) \sin (\frac{3 \pi}{2}+x)} = $

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