Chapter 5, Problem 8RE

To determine

To Simplify: The trigonometric expression $\frac{\tan \theta }{1-{\cos }^2\theta }$

Answer to Problem 8RE

  $\frac{\tan \theta }{1-{\cos }^2\theta }=\sec \theta \csc \theta$

Explanation of Solution

Given information:

  $\frac{\tan \theta }{1-{\cos }^2\theta }$

Concept and Formula Used:

  $\sec \theta =\frac{1}{\cos \theta },\csc \theta =\frac{1}{\sin \theta },1-{\cos }^{2}x={\sin }^{2}x,\tan \theta =\frac{\sin \theta }{\cos \theta }$

Calculation:

The given trigonometric expression is

  $\frac{\tan \theta }{1-{\cos }^2\theta }$

As

  $\begin{array}{l}1-{\cos }^{2}x={\sin }^{2}x\\ \frac{\tan \theta }{1-{\cos }^2\theta }=\frac{\tan \theta }{{\sin }^2\theta }\\ \text{Also, }\tan \theta =\frac{\sin \theta }{\cos \theta }\\ \frac{\tan \theta }{1-{\cos }^2\theta }=\frac{\sin \theta }{\cos \theta }\times \frac{1}{{\sin }^2\theta }\\ \frac{\tan \theta }{1-{\cos }^2\theta }=\frac{1}{\cos \theta \sin \theta }\\ \text{As sec}\theta =\frac{1}{\cos \theta },\csc \theta =\frac{1}{\sin \theta }\\ \frac{\tan \theta }{1-{\cos }^2\theta }=\sec \theta \csc \theta \end{array}$

Conclusion:

  $\frac{\tan \theta }{1-{\cos }^2\theta }=\sec \theta \csc \theta$