Chapter 5.2, Problem 31E

To determine

Toverify:The identity $\frac{{\cot }^{2}t}{\csc t}=\frac{1-{\sin }^{2}t}{\sin t}$ by converting the left side into sines and cosines.

Explanation of Solution

Given information:

The given identity is $\frac{{\cot }^{2}t}{\csc t}=\frac{1-{\sin }^{2}t}{\sin t}$ .

Formula used:

Quotient identity:

  ${\cot }^{2}t=\frac{{\cos }^{2}t}{{\sin }^{2}t}$

Reciprocal identity:

  $\csc t=\frac{1}{\sin t}$

Identity:

  ${\cos }^{2}t=1-{\sin }^{2}t$

Proof:

Let’s take left hand side of the identity.

  $\begin{array}{c}\frac{{\cot }^{2}t}{\csc t}=\frac{\frac{{\cos }^{2}t}{{\sin }^{2}t}}{\frac{1}{\sin t}}\left\{\begin{array}{l}\text{Quotient}\text{and}\text{reciprocal}\text{identity}\\ {\cot }^{2}t=\frac{{\cos }^{2}t}{{\sin }^{2}t}\\ \csc t=\frac{1}{\sin t}\end{array}\right\}\\ =\frac{{\cos }^{2}t}{{\sin }^{\cancel{2}}t}\times \cancel{\sin t}\\ =\frac{{\cos }^{2}t}{\sin t}\end{array}$

Further simplify the above expression.

  $\begin{array}{c}\frac{{\cot }^{2}t}{\csc t}=\frac{{\cos }^{2}t}{\sin t}\left\{{\cos }^{2}t=1-{\sin }^{2}t\right\}\\ =\frac{1-{\sin }^{2}t}{\sin t}\end{array}$

From the above, it is clear that the left hand side of the identity is equal to the right hand side of the identity.

Hence, proved.