Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sec \left(\frac{3\mathrm{\pi }}{2}-x\right)\sec \left(x-\frac{5\mathrm{\pi }}{2}\right)+\tan \left(\frac{5\mathrm{\pi }}{2}+x\right)\tan \left(x-\frac{3\mathrm{\pi }}{2}\right) \\ =\sec \left(\frac{3\mathrm{\pi }}{2}-x\right)\sec \left[-\left(\frac{5\mathrm{\pi }}{2}-x\right)\right]+\tan \left(\frac{5\mathrm{\pi }}{2}+x\right)\tan \left[-\left(\frac{3\mathrm{\pi }}{2}-x\right)\right] \\ =\sec \left(\frac{3\mathrm{\pi }}{2}-x\right)\sec \left(\frac{5\mathrm{\pi }}{2}-x\right)+\tan \left(\frac{5\mathrm{\pi }}{2}+x\right)\left[-\tan \left(\frac{3\mathrm{\pi }}{2}-x\right)\right] \\ =\sec \left(\frac{3\mathrm{\pi }}{2}-x\right)\sec \left(\frac{5\mathrm{\pi }}{2}-x\right)-\tan \left(\frac{5\mathrm{\pi }}{2}+x\right)\tan \left(\frac{3\mathrm{\pi }}{2}-x\right) \\ =\sec \left(\frac{\mathrm{\pi }}{2}\times 3-x\right)\sec \left(\frac{\mathrm{\pi }}{2}\times 5-x\right)-\tan \left(\frac{\mathrm{\pi }}{2}\times 5+x\right)\tan \left(\frac{\mathrm{\pi }}{2}\times 3-x\right) \\ =\left[-\mathrm{cosec}x\right]\left[\mathrm{cosec}x\right]-\left[-\cot x\right]\cot x \\ =-{\mathrm{cosec}}^{2}x+{\cot }^{2}x \\ =-\left[{\mathrm{cosec}}^{2}x-{\cot }^{2}x\right] \\ =-1 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$