Chapter 5, Problem 24RE

To determine

To simplify: the expression, check result numerically.

Answer to Problem 24RE

  $\frac{{\sin }^2\alpha -{\cos }^2\alpha }{{\sin }^2\alpha -\sin \alpha \cos \alpha }=1+\cot \alpha$ .

Explanation of Solution

Given information:

Given expression

  $\frac{{\sin }^2\alpha -{\cos }^2\alpha }{{\sin }^2\alpha -\sin \alpha \cos \alpha }$ .

Calculation:

Rewrite the number of the given expression.

  $\frac{{\sin }^2\alpha -{\cos }^2\alpha }{{\sin }^2\alpha -\sin \alpha \cos \alpha }=\frac{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}{{\sin }^2\alpha -\sin \alpha \cos \alpha }$

Factor out $\sin \alpha$ from the denominator of the expression.

  $\frac{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}{{\sin }^2\alpha -\sin \alpha \cos \alpha }=\frac{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}{\sin \alpha \left(\sin \alpha -\cos \alpha \right)}$

Simplify,

  $\begin{array}{c}\frac{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}{\sin \alpha \left(\sin \alpha -\cos \alpha \right)}=\frac{\sin \alpha +\cos \alpha }{\sin \alpha }\\ =\frac{\sin \alpha }{\sin \alpha }+\frac{\cos \alpha }{\sin \alpha }\\ =1+\frac{\cos \alpha }{\sin \alpha }\end{array}$

By the quotient identity, $\cot \alpha =\frac{\cos \alpha }{\sin \alpha }$

  $1+\frac{\cos \alpha }{\sin \alpha }=1+\cot \alpha$

Therefore, the expression simplifies to $1+\cot \alpha$ .

Now, check the result graphically as shown:

Table of values for $\frac{{\sin }^2\alpha -{\cos }^2\alpha }{{\sin }^2\alpha -\sin \alpha \cos \alpha }$ is shown:

  

Graph of $\frac{{\sin }^2\alpha -{\cos }^2\alpha }{{\sin }^2\alpha -\sin \alpha \cos \alpha }$ is constructed as

  

Table of values for $1+\cot \alpha$ is shown:

  

Graph of $1+\cot \alpha$ is constructed as shown:

  

Both graphs are same and the functional values are also same.