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If \(x \neq 0\), then \(\frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \operatorname{cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)}=\)
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Verified Answer
The correct answer is:
1
\(\begin{aligned}
& \frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \operatorname{cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)} \\
& (x \neq 0) \\
& =\frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x)}=1
\end{aligned}\)
Hence, option (c) is correct.
& \frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \operatorname{cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)} \\
& (x \neq 0) \\
& =\frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x)}=1
\end{aligned}\)
Hence, option (c) is correct.
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