Chapter 4.3, Problem 63E

To determine

To prove:The equation $\left(1+\cos \theta \right)\left(1-\cos \theta \right)={\sin }^2\theta$ when $0<\theta <\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The given equationis $\left(1+\cos \theta \right)\left(1-\cos \theta \right)={\sin }^2\theta$ .

Formula Used:

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

  $\left(a-b\right)\left(a+b\right)=a^2-b^2$

Proof:

Consider the left hand side of the equation and simplify it.

  $\begin{array}{c}\left(1+\cos \theta \right)\left(1-\cos \theta \right)=\left(1^2-{\cos }^2\theta \right)\left\{\left(a-b\right)\left(a+b\right)=a^2-b^2\right\}\\ =1-{\cos }^2\theta \\ ={\sin }^2\theta +\cancel{{\cos }^2\theta }-\cancel{{\cos }^2\theta }\left\{{\sin }^2\theta +{\cos }^2\theta =1\right\}\\ ={\sin }^2\theta \end{array}$

From the above, it is clear that one side of the equation can be transformed into other side of the equation.

Hence, proved.