Question

Prove that:
tan^2A +cot^2A+2=sec^2A *cosec^2A

Answer 1

$$\begin{gathered} \mathrm{LHS}={\tan }^2\mathrm{A}+{\cot }^2\mathrm{A}+2 \\ =\left({\tan }^2\mathrm{A}+1\right)+\left({\cot }^2\mathrm{A}+1\right) \\ ={\sec }^2\mathrm{A}+{\mathrm{cosec}}^2\mathrm{A} \\ =\frac{1}{{\cos }^2\mathrm{A}}+\frac{1}{{\sin }^2\mathrm{A}} \\ =\frac{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}.{\cos }^2\mathrm{A}} \\ =\frac{1}{{\sin }^2\mathrm{A}.{\cos }^2\mathrm{A}} \\ =\frac{1}{{\sin }^2\mathrm{A}}.\frac{1}{{\cos }^2\mathrm{A}} \\ ={\mathrm{cosec}}^2\mathrm{A}.{\sec }^2\mathrm{A} \\ =\mathrm{RHS} \end{gathered}$$