Chapter 5.2, Problem 26E

To determine

To verify: The given trigonometric identity algebraically and graphically.

Explanation of Solution

Given information:

The trigonometric identity ${\cot }^{2}y\left({\sec }^{2}y-1\right)=1$ .

Formula used:

The Pythagorean identity, ${\sin }^{2}x+{\cos }^{2}x=1$ .

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

Calculation:

Consider the function, ${\cot }^{2}y\left({\sec }^{2}y-1\right)=1$ .

The left hand side of equation is, ${\cot }^{2}y\left({\sec }^{2}y-1\right)$ .

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

Apply it in the left hand side of the equation.

  $\begin{array}{c}{\cot }^{2}y\left({\sec }^{2}y-1\right)=\frac{{\cos }^{2}y}{{\sin }^{2}y}\left(\frac{1}{{\cos }^{2}y}-1\right)\\ =\frac{{\cos }^{2}y}{{\sin }^{2}y}\left(\frac{1-{\cos }^{2}y}{{\cos }^{2}y}\right)\end{array}$

Recall Pythagorean identity, ${\sin }^{2}x+{\cos }^{2}x=1$ .

Apply it,

  $\begin{array}{c}{\cot }^{2}y\left({\sec }^{2}y-1\right)=\frac{{\cos }^{2}y}{{\sin }^{2}y}\left(\frac{1}{{\cos }^{2}y}-1\right)\\ =\frac{{\cos }^{2}y}{{\sin }^{2}y}\left(\frac{1-{\cos }^{2}y}{{\cos }^{2}y}\right)\\ =\frac{{\cos }^{2}y}{{\sin }^{2}y}\left(\frac{{\sin }^{2}y}{{\cos }^{2}y}\right)\\ =1\end{array}$

Since, left hand side and right hand side is equal, therefore, the trigonometric identity ${\cot }^{2}y\left({\sec }^{2}y-1\right)=1$ is verified.

To verify the result graphically, construct the graphs of the equations $y_1={\cot }^{2}x\left({\sec }^{2}x-1\right)$ and $y_2=1$ on the same window.

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

Step 2. Enter the function $\left(\frac{\text{COS}\left(X\right)}{\text{SIN}\left(X\right)}\right)^2\left(\frac{1}{{\left(\text{COS}\left(X\right)\right)}^2}-1\right)$ as $\backslash {\text{Y}}_1=$ .

Step 3. Enter the function $1$ as $\backslash {\text{Y}}_2=$ .

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

The result obtained is provided below,

  

Observe that graph of $y_1={\cot }^{2}x\left({\sec }^{2}x-1\right)$ and $y_2=1$ coincide with each other.

Therefore, the identity holds true.

Hence, the trigonometric identity ${\cot }^{2}x\left({\sec }^{2}x-1\right)=1$ is verified both algebraically and graphically