Prove That: Sec ( 3 π 2 − X ) Sec ( X − 5 π 2 ) + Tan ( 5 π 2 + X ) Tan ( X − 3 π 2 ) = − 1

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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Trigonometric Equations

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Q 5 Q 4 Q 6.1

पाठ 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ४०]

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आर.डी. शर्मा Mathematics [English] Class 11

पाठ 5 Trigonometric Functions
Exercise 5.3 | Q 5 | पृष्ठ ४०

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