Question

Find the value of: ${\sec }^2\theta -{\tan }^2\theta$

Answer 1

${\sec }^2\theta -{\tan }^2\theta =\frac{1}{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{1-{\sin }^2\theta }{{\cos }^2\theta }=\frac{{\cos }^2\theta }{{\cos }^2\theta }=1$

Since ${\sin }^2\theta +{\cos }^2\theta =1$