Chapter 13.6, Problem 46WE

To determine

To simplify: The each expression.

Answer to Problem 46WE

The simplified form of expression is $2\csc \alpha$ .

Explanation of Solution

Given:

The given expression is as:

  $\frac{\sin \alpha }{1+\cos \alpha }+\frac{1+\cos \alpha }{\sin \alpha }$ .

The simplified form is calculated as:

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

Further simplified as:

  $\begin{array}{c}\frac{\sin \alpha }{1+\cos \alpha }+\frac{1+\cos \alpha }{\sin \alpha }=\frac{1-\cos \alpha +1+\cos \alpha }{\sin \alpha }\\ =\frac{2}{\sin \alpha }\\ =2\csc \alpha \end{array}$

Therefore the simplified form of expression is $2\csc \alpha$ .