Chapter 5.2, Problem 66E

To determine

To explain: The given trigonometric identity with help of the figure.

Explanation of Solution

Given information:

The given trigonometric identity is $\frac{{\sec }^2\theta -1}{{\sec }^2\theta }={\sin }^2\theta$

Formula used:

Trigonometric function, $\sec x=\frac{\text{hypotenuse}}{\text{base}}$ and $\sin x=\frac{\text{perpendicular}}{\text{hypotenuse}}$ .

Calculation:

Consider the function, $\frac{{\sec }^2\theta -1}{{\sec }^2\theta }={\sin }^2\theta$ .

Recall that $\sec x=\frac{\text{hypotenuse}}{\text{base}}$ and $\sin x=\frac{\text{perpendicular}}{\text{hypotenuse}}$ .

From the figure

  

Hypotenuse is c, base is b and perpendicular is a.

In a right angle triangle $c^2=a^2+b^2$

Rewrite the left hand side of the equation as,

  $\begin{array}{c}\frac{{\sec }^2\theta -1}{{\sec }^2\theta }=\frac{{\left(\frac{c}{b}\right)}^2-1}{{\left(\frac{c}{b}\right)}^2}\\ =\frac{\frac{c^2-b^2}{b^2}}{\frac{c^2}{b^2}}\\ =\frac{c^2-b^2}{c^2}\\ =\frac{a^2}{c^2}\end{array}$

Rewrite the right hand side of the equation as,

  ${\sin }^2\theta =\frac{a^2}{c^2}$

Thus, left hand side and right hand side is equal, therefore, the trigonometric identity $\frac{{\sec }^2\theta -1}{{\sec }^2\theta }={\sin }^2\theta$ is verified.