Chapter 7.4, Problem 62AYU

establish each identity.

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

To determine

To find: The trigonometric expression $\frac{\sec \theta - \cos \theta }{\sec \theta +\cos \theta }= \frac{{\text{sin}}^2\theta }{1+ {\text{cos}}^2\theta }$.

Answer to Problem 62AYU

Solution:

The trigonometric expression $\frac{\sec \theta - \cos \theta }{\sec \theta +\cos \theta }= \frac{{\text{sin}}^2\theta }{1+ {\text{cos}}^2\theta }$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{\sec \theta - \cos \theta }{\sec \theta +\cos \theta }= \frac{{\text{sin}}^2\theta }{1+ {\text{cos}}^2\theta }$.

Calculation:

Left hand side,

$\frac{\sec \theta - \cos \theta }{\sec \theta +\cos \theta }$

By quotient identity for $\sec \theta$.

$\frac{\frac{1}{\cos \theta }- \cos \theta }{\frac{1}{\cos \theta } +\cos \theta }$

Adding up the quotients in the numerator and denominator.

$\frac{\frac{1-{\text{ cos}}^2\theta }{\cos \theta }}{\frac{1+ {\text{cos}}^2\theta }{\cos \theta }}$

By Pythagorean identity,

$\frac{\frac{{\text{sin}}^2\theta }{\cos \theta }}{\frac{1+ {\text{cos}}^2\theta }{\cos \theta }}$

Cancelling the common factor $\text{cos}\theta$ by reciprocal.

$\frac{{\text{sin}}^2\theta }{1+ {\text{cos}}^2\theta }$

$\text{LHS }=\text{ RHS}$