Chapter 5.2, Problem 62E

To determine

To graph: The given trigonometric function and make a conjecture about the simplification of the expression. Then verify the resulting identity algebraically.

Explanation of Solution

Given information:

The following trigonometric function:

  $y=\frac{\cos x}{1-\tan x}+\frac{\sin x\cos x}{\sin x-\cos x}$

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\frac{\cos x}{1-\tan x}+\frac{\sin x\cos x}{\sin x-\cos x}$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

Interpretation:

According to the above graph, it can be observed that $y=\cos x$

Formula used:

The following identity:

  $\tan x=\frac{\sin x}{\cos x}$

Proof:

The given trigonometric function can be simplified as follows:

  $\begin{array}{l}y=\frac{\cos x}{1-\tan x}+\frac{\sin x\cos x}{\sin x-\cos x}\text{ [Original expression]}\\ =\frac{\cos x}{1-\frac{\sin x}{\cos x}}\text{+ }\frac{\sin x\cos x}{\sin x-\cos x}\text{ [Quotient identity]}\\ =\frac{\cos x}{\frac{\cos x-\sin x}{\cos x}}\text{+}\frac{\sin x\cos x}{\sin x-\cos x}\text{ [Subtract the fraction in the denominator]}\\ \text{ =}\frac{{\cos }^{2}x}{\cos x-\sin x}\text{+ }\frac{\sin x\cos x}{\sin x-\cos x}\text{ [Simplify]}\\ =\frac{{\cos }^{2}x}{\cos x-\sin x}-\frac{\sin x\cos x}{\cos x-\sin x}\text{ [Change sign to make denominators equal]}\\ =\frac{{\cos }^{2}x-\sin x\cos x}{\cos x-\sin x}\text{ [Subtract fractions]}\\ =\frac{\cos x(\cos x-\sin x)}{\cos x-\sin x}\text{ [Factor]}\\ \text{ = cos }x\text{ [Divide]}\end{array}$

Therefore, $\text{y}=\cos x$ .

The simplified expression is same as the expression obtained from the graph. Hence, verified.