Question

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

Answer 1

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