Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Summary Exercises on Verifying Trigonometric Identities - Page 245: 5


$$\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$$ As in the Work Step by Step, prove that 2 sides are equal and it follows that the equation is an identity.

Work Step by Step

$$\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$$ You can choose any side to start with in this exercise. I choose the left one to start with here. $$X=\frac{\sin t}{1+\cos t}$$ The trick here is to multiply both numerator and denominator by $1-\cos t$. $$X=\frac{\sin t(1-\cos t)}{(1+\cos t)(1-\cos t)}$$ $$X=\frac{\sin t(1-\cos t)}{1-\cos^2t}$$ (since $(A-B)(A+B)=A^2-B^2$) - Now, according to Pythagorean identity: $\sin^2t=1-\cos^2t$ $$X=\frac{\sin t(1-\cos t)}{\sin^2t}$$ $$X=\frac{1-\cos t}{\sin t}$$ Therefore, $$\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$$ So the equation is an identity.
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