Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 14


$\frac{tan~x-cot~x}{tan~x+cot~x} = 2sin^2~x-1$

Work Step by Step

$\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{\frac{sin~x}{cos~x}-\frac{cos~x}{sin~x}}{\frac{sin~x}{cos~x}+\frac{cos~x}{sin~x}}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{\frac{sin^2~x-cos^2~x}{sin~x~cos~x}}{\frac{sin^2~x+cos^2~x}{sin~x~cos~x}}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{sin^2~x-cos^2~x}{sin^2~x+cos^2~x}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{sin^2~x-cos^2~x}{1}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = sin^2~x-(1-sin^2~x)$ $\frac{tan~x-cot~x}{tan~x+cot~x} = 2sin^2~x-1$
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