Chapter 5.2, Problem 29E

To determine

To verify: The given trigonometric identity algebraically and graphically.

Explanation of Solution

Given information:

The trigonometric identity $\frac{1}{\tan \beta }+\tan \beta =\frac{{\sec }^2\beta }{\tan \beta }$ .

Formula used:

The Pythagorean identity $1+{\tan }^{2}x={\sec }^{2}x$ .

Calculation:

Consider the function, $\frac{1}{\tan \beta }+\tan \beta =\frac{{\sec }^2\beta }{\tan \beta }$ .

The left hand side of equation is, $\frac{1}{\tan \beta }+\tan \beta$ .

  $\frac{1}{\tan \beta }+\tan \beta =\frac{1+{\tan }^2\beta }{\tan \beta }$

Recall Pythagorean identity $1+{\tan }^{2}x={\sec }^{2}x$ .

Apply it in the left hand side of the equation.

  $\begin{array}{c}\frac{1}{\tan \beta }+\tan \beta =\frac{1+{\tan }^2\beta }{\tan \beta }\\ =\frac{{\sec }^2\beta }{\tan \beta }\end{array}$

Since, left hand side and right hand side is equal, therefore, the trigonometric identity $\frac{1}{\tan \beta }+\tan \beta =\frac{{\sec }^2\beta }{\tan \beta }$ is verified.

To verify the result graphically, construct the graphs of the equations $y_1=\frac{1}{\tan x}+\tan x$ and $y_2=\frac{{\sec }^{2}x}{\tan x}$ on the same window.

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

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

Step 3. Enter the function $\left(\left(1÷\cos \left(x\right)\right)^2\right)÷\tan \left(x\right)$ as $\backslash {\text{Y}}_2=$ .

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

The result obtained is provided below,

  

Observe that graph of $y_1=\frac{1}{\tan x}+\tan x$ and $y_2=\frac{{\sec }^{2}x}{\tan x}$ coincide with each other.

Therefore, the identity holds true.

Hence, the trigonometric identity $\frac{1}{\tan \beta }+\tan \beta =\frac{{\sec }^2\beta }{\tan \beta }$ is verified both algebraically and graphically