Chapter 5.2, Problem 49E

To determine

To verify the identity:

  $\frac{{\sin }^2\theta }{1-\cos \theta }=\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }$

And also verify this using table feature of graphing utility.

Explanation of Solution

Given:

  $\frac{{\sin }^2\theta }{1-\cos \theta }=\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }$

Concept Used:

The trigonometric identities:

 • $\sec x=\frac{1}{\cos x}$

 • ${\sin }^{2}x+{\cos }^{2}x=1$

Calculation:

In order to verify the identity:

  $\frac{{\sin }^2\theta }{1-\cos \theta }=\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

  $\begin{array}{c}\frac{{\sin }^2\theta }{1-\cos \theta }=\frac{{\sin }^2\theta \left(1+\cos \theta \right)}{\left(1-\cos \theta \right)\left(1+\cos \theta \right)}\\ =\frac{\left(1-{\cos }^2\theta \right)\left(1+\cos \theta \right)}{\left(1-\cos \theta \right)\left(1+\cos \theta \right)}\\ =\frac{\left(1-\cos \theta \right){\left(1+\cos \theta \right)}^2}{\left(1-\cos \theta \right)\left(1+\cos \theta \right)}\\ =\frac{1+{\cos }^2\theta +2\cos \theta }{\left(1+\cos \theta \right)}\\ =\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{{\sin }^2\theta }{1-\cos \theta }=\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }$

Table of values for $\frac{{\sin }^2\theta }{1-\cos \theta }$ is shown:

  

Graph of $\frac{{\sin }^2\theta }{1-\cos \theta }$ is constructed as

  

Table of values for $\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }$ is shown:

  

Graph of $\frac{2+2\cos \theta -{\sin }^2\theta }{1+\cos \theta }$ is constructed as shown:

  

Since, the graph of left hand side and right hand side of the identity coincides, thus the identity is verified by graph too.