Chapter 7, Problem 18SGA

To determine

To verify: that the given equation is an identity.

Answer to Problem 18SGA

See explanation

Explanation of Solution

Given information:

  $\frac{\sec \theta +1}{\tan \theta }=\frac{\tan \theta }{\sec \theta -1}$

Calculation:

Simplify the left side of the equation as follows:

  $\begin{array}{c}\frac{\sec \theta +1}{\tan \theta }=\frac{\left(\frac{1}{\cos \theta }+1\right)}{\left(\frac{\sin \theta }{\cos \theta }\right)}\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta }\right]\end{array}$

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

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

  $\begin{array}{c}=\left(\frac{1+\cos \theta }{\sin \theta }\right)\left(\frac{1-\cos \theta }{1-\cos \theta }\right)\left[\begin{array}{l}\text{Multiply numerator and }\\ \text{denominator by }\left(1-\cos \theta \right)\end{array}\right]\end{array}$

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

  $\begin{array}{c}=\frac{{\sin }^2\theta }{\sin \theta \left(1-\cos \theta \right)}\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\end{array}$

  $\begin{array}{c}=\frac{\sin \theta }{1-\cos \theta }\end{array}$

Now simplify the right side of the equation as follows:

  $\begin{array}{c}\frac{\tan \theta }{\sec \theta -1}=\frac{\left(\frac{\sin \theta }{\cos \theta }\right)}{\left(\frac{1}{\cos \theta }-1\right)}\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta }\text{ and }\sec \theta =\frac{1}{\cos \theta }\right]\end{array}$

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

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

  $\begin{array}{c}=LHS\end{array}$

Since left hand side and right hand side of the equation equal. So, the equation is an identity.