Chapter 5, Problem 24RE

Solution

To find: The given equation $\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\tan }^2\alpha +1\right)={\tan }^2\alpha -1$ is likely to be an identity.

It is proved that the given equation $\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\tan }^2\alpha +1\right)={\tan }^2\alpha -1$ is likely to be an identity.

Given information:

The given equation is $\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\tan }^2\alpha +1\right)={\tan }^2\alpha -1$ .

Formula used:

The formula of Pythagorean identity is $1+{\tan }^2\theta ={\sec }^2\theta$ .

The formula of basic identities of trigonometric functions are $\sec \theta =\frac{1}{\cos \theta }$ and $\tan \theta =\frac{\sin \theta }{\cos \theta }$ .

Calculation:

Apply the formula of Pythagorean identity, ${\tan }^2\theta +1={\sec }^2\theta$ and basic identities of trigonometric functions, $\sec \theta =\frac{1}{\cos \theta }$ , $\tan \theta =\frac{\sin \theta }{\cos \theta }$ into left side of the given equation and simplify.

  $\begin{array}{c}\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\tan }^2\alpha +1\right)=\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\sec }^2\alpha \right)\\ =\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left(\frac{1}{{\cos }^2\alpha }\right)\\ =\frac{{\sin }^2\alpha -{\cos }^2\alpha }{{\cos }^2\alpha }\\ =\frac{{\sin }^2\alpha }{{\cos }^2\alpha }-\frac{{\cos }^2\alpha }{{\cos }^2\alpha }\\ ={\tan }^2\alpha -1\end{array}$

The obtained expression is same as the given right side expression. So it is proved that the given equation is likely to be an identity.

Draw the graph of the functions $y=\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\tan }^2\alpha +1\right)$ and $y={\tan }^2\alpha -1$ on the graphing calculator.

The graph of the function $y=\left({\sin }^2\alpha -{\cos }^2\alpha \right)\left({\tan }^2\alpha +1\right)$ is shown below.

  

The graph of the function $y={\tan }^2\alpha -1$ is shown below.

  

Both the obtained graphs are similar and appear to be identical.

Thus, it can be said that the given equation is likely to be an identity from graphical support also.