Question

Prove that $\left(1+\frac{1}{{\tan }^{2}A}\right)\times \left(1+\frac{1}{{\cot }^{2}A}\right)=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}$

Answer 1

Consider the problem

$\begin{array}{c}\left(1+\frac{1}{{\tan }^{2}A}\right)\times \left(1+\frac{1}{{\cot }^{2}A}\right)\\ =\left(1+{\cot }^{2}A\right)\left(1+{\tan }^{2}A\right)\left(\because \tan A=\frac{1}{\cot A}\right)\\ =\cos e{c}^{2}A{\sec }^{2}A\left(\begin{array}{c}\because \cos e{c}^{2}A=1+{\cot }^{2}A\\ {\sec }^{2}A=1+{\tan }^{2}A\end{array}\right)\\ =\frac{1}{{\sin }^{2}A}\times \frac{1}{{\cos }^{2}A}\left(\because sinA=\frac{1}{\cos ecA},\sec A=\frac{1}{\cos A}\right)\\ =\frac{1}{{\sin }^{2}A\left(1-{\sin }^{2}A\right)}\left(\because {\sin }^{2}A+{\cos }^{2}A=1\right)\\ =\frac{1}{{\sin }^{2}A-{\sin }^{4}A}\end{array}$