Chapter 13.1, Problem 2AGP

To determine

The simplified form of expression.

Answer to Problem 2AGP

  ${\sin }^2\theta$

Explanation of Solution

Given:

The given expression is

  $\frac{{\tan }^2\theta {\csc }^2\theta -1}{{\sec }^2\theta }$

Calculation:

Solving the expression as follows-

  $\begin{array}{l}\frac{{\tan }^2\theta {\csc }^2\theta -1}{{\sec }^2\theta }\\ =\frac{\frac{{\sin }^2\theta }{{\cos }^2\theta }\cdot \frac{1}{{\sin }^2\theta }-1}{\frac{1}{{\cos }^2\theta }}\\ =\frac{\frac{1}{{\cos }^2\theta }-1}{\frac{1}{{\cos }^2\theta }}\\ =\frac{\frac{1-{\cos }^2\theta }{{\cos }^2\theta }}{\frac{1}{{\cos }^2\theta }}\\ =1-{\cos }^2\theta \\ ={\sin }^2\theta \end{array}$

The simplest expression is ${\sin }^2\theta$