Question

Find the value of $9{\sec }^{2}A-9{\tan }^{2}A$

Answer 1

$\begin{array}{c}9{\sec }^{2}A-9{\tan }^{2}A,\left[1-{\sin }^{2}A={\cos }^{2}A\right]\\ \Rightarrow \frac{9}{{\cos }^{2}A}-\frac{9{\sin }^{2}A}{{\cos }^{2}A}=\frac{9\left(1-{\sin }^{2}A\right)}{{\cos }^{2}A}=\frac{9{\cos }^{2}A}{{\cos }^{2}A}\\ =9Ans.\end{array}$