Question

(sec A/cotA+tanA)^2​

Answer 1

Answer:

$lhs= ( \frac{ \sec \alpha }{ \cot\alpha + \tan\alpha } ) {}^{2} \\ lhs = \frac{ \sec {}^{2} \alpha }{ \cot {}^{2} \alpha + \tan {}^{2} \alpha } \\ lhs = \frac{\sec {}^{2} \alpha}{ \frac{ + \frac{?}{?} \sin {}^{2} \alpha }{ \cos {}^{2} \alpha } }$