## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 243: 57

#### Answer

$\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}}= sec~x$

#### Work Step by Step

$\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}}$ When we graph this function, we can see that it looks like the graph of $~~sec~x$ Note that: $~cos~2a = cos^2~a-sin^2~a$ We can verify this algebraically: $\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}} = \frac{\frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}+\frac{cos~\frac{x}{2}}{sin~\frac{x}{2}}}{\frac{cos~\frac{x}{2}}{sin~\frac{x}{2}}-\frac{sin~\frac{x}{2}}{cos~\frac{x}{2}}}$ $\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}} = \frac{\frac{sin^2~\frac{x}{2}+~cos^2~\frac{x}{2}}{cos~\frac{x}{2}~sin~\frac{x}{2}}}{\frac{cos^2~\frac{x}{2}-~sin^2~\frac{x}{2}}{sin~\frac{x}{2}~cos~\frac{x}{2}}}$ $\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}} = \frac{sin^2~\frac{x}{2}+~cos^2~\frac{x}{2}}{cos^2~\frac{x}{2}-~sin^2~\frac{x}{2}}$ $\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}} = \frac{1}{cos^2~\frac{x}{2}-~sin^2~\frac{x}{2}}$ $\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}} = \frac{1}{cos~x}$ $\frac{tan~\frac{x}{2}+cot~\frac{x}{2}}{cot~\frac{x}{2}-tan~\frac{x}{2}}= sec~x$

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