Chapter 5.1, Problem 50E

Solution

To find: The algebraic expression with single trigonometric function for $\frac{{\tan }^{2}x}{\sec x+1}$ .

The algebraic expression with single trigonometric function for $\frac{{\tan }^{2}x}{\sec x+1}$ is $\left(\sec x-1\right)$ .

Given information:

The given expression is $\frac{{\tan }^{2}x}{\sec x+1}$ .

Formula used:

The Pythagorean Identities for ${\tan }^{2}x$ and ${\sec }^{2}x$ is given by,

  $\begin{array}{c}{\sec }^{2}x-{\tan }^{2}x=1\\ {\tan }^{2}x={\sec }^{2}x-1\end{array}$

The formula for the difference of squares of two terms is given by,

  $a^2-b^2=\left(a+b\right)\left(a-b\right)$

Calculation:

Consider the given expression,

  $\Rightarrow \frac{{\tan }^{2}x}{\sec x+1}$

Use the Pythagorean identity and substitute ${\sec }^{2}x-1$ for ${\tan }^{2}x$ in the expression.

  $\Rightarrow \frac{{\tan }^{2}x}{\sec x+1}=\frac{{\sec }^{2}x-1}{\sec x+1}$

Use the identity for the difference of squares of two terms to simplify the numerator of the expression.

  $\Rightarrow \frac{\left(\sec x-1\right)\left(\sec x+1\right)}{\sec x+1}$

Cancel the similar terms, $\left(\sec x+1\right)$ .

  $\Rightarrow \left(\sec x-1\right)$

Thus, the algebraic expression with single trigonometric function for $\frac{{\tan }^{2}x}{\sec x+1}$ is $\left(\sec x-1\right)$ .