Chapter 5.2, Problem 70E

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:

  $y=\sin t+\frac{{\cot }^{2}t}{\csc t}$

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 $\sin t+\frac{{\cot }^{2}t}{\csc t}$ 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 $y=\csc t$

Formula used:

The following identities:

  $\begin{array}{l}\cot x=\frac{\cos x}{\sin x}\\ \csc x=\frac{1}{\sin x}\\ {\sin }^{2}x+{\cos }^{2}x=1\end{array}$

Proof:

The given trigonometric function can be simplified as follows:

  $\begin{array}{l}y=\sin t+\frac{{\cot }^{2}t}{\csc t}\text{ [Original expression]}\\ =\sin t+\frac{\frac{{\cos }^{2}t}{{\sin }^{2}t}}{\frac{1}{\sin t}}\text{ [Quotient identity and reciprocal identity]}\\ =\sin t+\frac{{\cos }^{2}t}{{\sin }^{2}t}\times \frac{\sin t}{1}\text{ [Multiply by reciprocal of denominator]}\\ \text{ = }\sin t+\frac{{\cos }^{2}t}{\sin t}\text{ [Simplify]}\\ =\frac{{\sin }^{2}t+{\cos }^{2}t}{\sin t}\text{ [Add fractions]}\\ =\frac{1}{\sin t}\text{ [Pythagorean identity]}\\ =\csc t\text{ [Reciprocal identity]}\\ \end{array}$

Therefore, $\text{y}=\csc t$ .

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