Chapter 5.2, Problem 27E

To determine

To verify: The given trigonometric identity algebraically and graphically.

Explanation of Solution

Given information:

The trigonometric identity $\frac{{\tan }^2\theta }{\sec \theta }=\sin \theta \tan \theta$ .

Formula used:

Conversion rules, $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$ .

Calculation:

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

The left hand side of equation is, $\frac{{\tan }^2\theta }{\sec \theta }$ .

Recall that $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$ .

Apply it in the left hand side of the equation.

  $\begin{array}{c}\frac{{\tan }^2\theta }{\sec \theta }=\tan \theta \frac{\tan \theta }{\sec \theta }\\ =\tan \theta \frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}\\ =\tan \theta \times \frac{\sin \theta }{\cos \theta }\times \frac{1}{\cos \theta }\\ =\tan \theta \sin \theta \end{array}$

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

To verify the result graphically, construct the graphs of the equations $y_1=\frac{{\tan }^{2}x}{\sec x}$ and $y_2=\sin x\tan x$ on the same window.

Step 1. Press $\boxed{\text{Y}=}$ key.

Step 2. Enter the function ${\left(\tan \left(x\right)\right)}^2÷\left(1÷\cos \left(x\right)\right)$ as $\backslash {\text{Y}}_1=$ .

Step 3. Enter the function $\sin \left(x\right)\times \tan \left(x\right)$ as $\backslash {\text{Y}}_2=$ .

Step 4: Press the $\boxed{\text{GRAPH}}$ key.

The result obtained is provided below,

  

Observe that graph of $y_1=\frac{{\tan }^{2}x}{\sec x}$ and $y_2=\sin x\tan x$ coincide with each other.

Therefore, the identity holds true.

Hence, the trigonometric identity $\frac{{\tan }^2\theta }{\sec \theta }=\sin \theta \tan \theta$ is verified both algebraically and graphically