Chapter 5, Problem 15RE

To determine

To simplify: the expression, check result numerically.

Answer to Problem 15RE

  $\frac{1}{{\tan }^{2}x+1}={\cos }^{2}x$

Explanation of Solution

Given information:

Given expression

  $\frac{1}{{\tan }^{2}x+1}$ .

Calculation:

Simplify $\frac{1}{{\tan }^{2}x+1}$ by applying the Pythagorean identity ${\sec }^{2}u=1+{\tan }^{2}u$ therefore;

  $\frac{1}{{\tan }^{2}x+1}={\sec }^{2}x$

Apply reciprocal identity $\cos x=\frac{1}{\sec x}$ ;

  $\begin{array}{c}\frac{1}{{\sec }^{2}x}={\left(\cos x\right)}^2\\ ={\cos }^{2}x\end{array}$

Therefore $\frac{1}{{\tan }^{2}x+1}={\cos }^{2}x$

Verify the result by use of T1-83 calculator, Let;

  $y_1=\frac{1}{\left({\tan }^{2}x+1\right)}$

And,

  $y_2={\cos }^{2}x$

Enter the value of function $y_1$ in $Y1=$ the display is as shown below;

  

Enter the value of function $y_2$ in $Y2=$ the display is as shown below;

  

Click $2^{nd}$ and window for TBL and SET and set values, the display is as shown below;

  

Click $2^{nd}$ and GRAPH for TABLE, the display for table is as shown below;

  

As from the table, the values of $y_1$ and $y_2$ are same for all values of $x$ . Therefore the function $y_1$ and the function $y_2$ are equivalent that is;

  $\frac{1}{{\tan }^{2}x+1}={\cos }^{2}x$