Chapter 5.1, Problem 26E

Solution

To simplify: The expression $\frac{{\sec }^{2}u-{\tan }^{2}u}{{\cos }^{2}v+{\sin }^{2}v}$ is either a constant or basic trigonometric function.

The simplified form of expression $\frac{{\sec }^{2}u-{\tan }^{2}u}{{\cos }^{2}v+{\sin }^{2}v}$ is 1.

Given information: The given expression is $\frac{{\sec }^{2}u-{\tan }^{2}u}{{\cos }^{2}v+{\sin }^{2}v}$ .

Formula used: The Pythagorean identities of the trigonometric function is as follows:

  $\begin{array}{l}\left({\sin }^{2}x+{\cos }^{2}x\right)=1\\ \left({\sec }^{2}x-{\tan }^{2}x\right)=1\end{array}$

Calculation:

Apply Pythagorean identities to simplify the expression.

  $\begin{array}{c}\frac{{\sec }^{2}u-{\tan }^{2}u}{{\cos }^{2}v+{\sin }^{2}v}=\frac{1}{1}\\ =1\end{array}$

Draw the graph of is $\frac{{\sec }^{2}u-{\tan }^{2}u}{{\cos }^{2}v+{\sin }^{2}v}$ and $1$ .

  

Figure $\left(1\right)$

It can be observed that the graph of both function is the same.

Thus, the simplified form of expression $\frac{{\sec }^{2}u-{\tan }^{2}u}{{\cos }^{2}v+{\sin }^{2}v}$ is $1$ .