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प्रश्न
Prove that:
\[\frac{\cos (2\pi + x) cosec (2\pi + x) \tan (\pi/2 + x)}{\sec(\pi/2 + x)\cos x \cot(\pi + x)} = 1\]
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उत्तर
LHS = \[\frac{\cos\left( 2\pi + x \right) cosec\left( 2\pi + x \right) \tan\left( \frac{\pi}{2} + x \right)}{\sec \left( \frac{\pi}{2} + x \right) \cos x \cot \left( \pi + x \right)}\]
\[ = \frac{\cos x cosec x \left[ - \cot x \right]}{\left[ - cosec x \right]\cos x \cot x} \]
\[ = \frac{- \cos x cosec x \cot x}{- cosec x cos x \cot x}\]
\[ = 1\]
= RHS
Hence proved.
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