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# Prove that:

Question:

Prove that:

$\sec \left(\frac{3 \pi}{2}-x\right) \sec \left(x-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+x\right) \tan \left(x-\frac{3 \pi}{2}\right)=-1$

Solution:

$\mathrm{LHS}=\sec \left(\frac{3 \pi}{2}-x\right) \sec \left(x-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+x\right) \tan \left(x-\frac{3 \pi}{2}\right)$

$=\sec \left(\frac{3 \pi}{2}-x\right) \sec \left[-\left(\frac{5 \pi}{2}-x\right)\right]+\tan \left(\frac{5 \pi}{2}+x\right) \tan \left[-\left(\frac{3 \pi}{2}-x\right)\right]$

$=\sec \left(\frac{3 \pi}{2}-x\right) \sec \left(\frac{5 \pi}{2}-x\right)+\tan \left(\frac{5 \pi}{2}+x\right)\left[-\tan \left(\frac{3 \pi}{2}-x\right)\right]$

$=\sec \left(\frac{3 \pi}{2}-x\right) \sec \left(\frac{5 \pi}{2}-x\right)-\tan \left(\frac{5 \pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right)$

$=\sec \left(\frac{\pi}{2} \times 3-x\right) \sec \left(\frac{\pi}{2} \times 5-x\right)-\tan \left(\frac{\pi}{2} \times 5+x\right) \tan \left(\frac{\pi}{2} \times 3-x\right)$

$=[-\operatorname{cosec} x][\operatorname{cosec} x]-[-\cot x] \cot x$

$=-\operatorname{cosec}^{2} x+\cot ^{2} x$

$=-\left[\operatorname{cosec}^{2} x-\cot ^{2} x\right]$

= - 1

=RHS

Hence, proved.