Chapter 9.7, Problem 18E

To determine

Simplify the expression.

Answer to Problem 18E

The simplified form of expression $\frac{{\cos }^{2}x{\tan }^2\left(-x\right)-1}{{\cos }^{2}x}$ is $-1$ .

Explanation of Solution

Given information:

The expression:

  $\frac{{\cos }^{2}x{\tan }^2\left(-x\right)-1}{{\cos }^{2}x}$

Formula used:

  ${\sec }^2\theta -{\tan }^2\theta =1$

  $\frac{1}{\cos \theta }=\sec \theta$

  $\tan \left(-\theta \right)=-\tan \theta$

  $\frac{{\cos }^{2}x{\tan }^2\left(-x\right)-1}{{\cos }^{2}x}$

Expanding the fraction by writing denominator below each terms of numerator,

  $=\frac{{\cos }^{2}x{\tan }^2\left(-x\right)}{{\cos }^{2}x}-\frac{1}{{\cos }^{2}x}$

Cancelling out ${\cos }^{2}x$ from first function,

  $={\tan }^2\left(-x\right)-\frac{1}{{\cos }^{2}x}$

Applying the identity $\frac{1}{\cos \theta }=\sec \theta$ ,

  $\begin{array}{l}={\tan }^2\left(-x\right)-{\sec }^{2}x\\ \\ =\tan \left(-x\right)\cdot \tan \left(-x\right)-{\sec }^{2}x\end{array}$

Applying the identity $\tan \left(-\theta \right)=-\tan \theta$ ,

  $\begin{array}{l}=\left(-\tan x\right)\cdot \left(-\tan x\right)-{\sec }^{2}x\\ \\ ={\tan }^{2}x-{\sec }^{2}x\end{array}$

Taking negative sign common,

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

Applying the identity ${\sec }^2\theta -{\tan }^2\theta =1$ ,

  $=-1$

Cancelling out $\sin x$

The simplified form of expression $\frac{{\cos }^{2}x{\tan }^2\left(-x\right)-1}{{\cos }^{2}x}$ is $-1$ .