Chapter 14, Problem 22SGR

To determine

To verify :The value of ${\sin }^2\theta -1=\frac{{\sin }^2\theta }{1-{\sin }^2\theta }$

Explanation of Solution

Given information :

We know that, the given information is $LHS={\sec }^2\theta -1$

Formula Used :

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

Calculation :

The given equation is

  $\begin{array}{l}LHS={\sec }^2\theta -1\\ =\frac{1}{{\cos }^2\theta }-1,\left[\sec \theta =\frac{1}{\cos \theta }\right]\\ =\frac{1-{\cos }^2\theta }{{\cos }^2\theta }=\frac{{\sin }^2\theta }{{\cos }^2\theta }\\ =\frac{{\sin }^2\theta }{1-{\sin }^2\theta },[{\sin }^2\theta +{\cos }^2\theta =1]\\ =RHS\end{array}$