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प्रश्न
Prove that:
\[\sec\left( \frac{3\pi}{2} - x \right)\sec\left( x - \frac{5\pi}{2} \right) + \tan\left( \frac{5\pi}{2} + x \right)\tan\left( x - \frac{3\pi}{2} \right) = - 1 .\]
उत्तर
LHS = \[\sec\left( \frac{3\pi}{2} - x \right)\sec\left( x - \frac{5\pi}{2} \right) + \tan\left( \frac{5\pi}{2} + x \right)\tan\left( x - \frac{3\pi}{2} \right)\]
\[ = \sec\left( \frac{3\pi}{2} - x \right)\sec\left[ - \left( \frac{5\pi}{2} - x \right) \right] + \tan\left( \frac{5\pi}{2} + x \right)\tan\left[ - \left( \frac{3\pi}{2} - x \right) \right]\]
\[ = \sec\left( \frac{3\pi}{2} - x \right)\sec\left( \frac{5\pi}{2} - x \right) + \tan\left( \frac{5\pi}{2} + x \right)\left[ - \tan\left( \frac{3\pi}{2} - x \right) \right]\]
\[ = \sec\left( \frac{3\pi}{2} - x \right)\sec\left( \frac{5\pi}{2} - x \right) - \tan\left( \frac{5\pi}{2} + x \right)\tan\left( \frac{3\pi}{2} - x \right)\]
\[ = \sec\left( \frac{\pi}{2} \times 3 - x \right)\sec\left( \frac{\pi}{2} \times 5 - x \right) - \tan\left( \frac{\pi}{2} \times 5 + x \right)\tan\left( \frac{\pi}{2} \times 3 - x \right)\]
\[ = \left[ - cosec x \right]\left[ cosec x \right] - \left[ - \cot x \right]\cot x \]
\[ = - {cosec}^2 x + \cot^2 x\]
\[ = - \left[ {cosec}^2 x - \cot^2 x \right]\]
\[ = - 1\]
= RHS
Hence proved.
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