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Q. 80

Expert-verified
Found in: Page 475

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Establish each identity. $\frac{\mathrm{sin}\theta +\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\mathrm{cos}\theta }=sec\theta csc\theta$

The left side of the given equation has been simplified to bring the right side.

See the step by step solution

## Step 1. Given Information

The given equation is $\frac{\mathrm{sin}\theta +\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\mathrm{cos}\theta }=sec\theta csc\theta$

## Step 2. Calculation

Consider the left side of the given equation and simplify to bring the right side in order to establish the equation.

$\frac{\mathrm{sin}\theta +\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\mathrm{cos}\theta }=\frac{\mathrm{cos}\theta \left(\mathrm{sin}\theta +\mathrm{cos}\theta \right)-\mathrm{sin}\theta \left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)}{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\theta \mathrm{sin}\theta +{\mathrm{cos}}^{2}\theta -\mathrm{sin}\theta \mathrm{cos}\theta +{\mathrm{sin}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=sec\theta csc\theta$

Hence, Proved