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: 6


$$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$ The proof is in the Work Step by Step.

Work Step by Step

$$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$ We take a look at the right side first. $$X=\frac{1}{\sec t+\tan t}$$ $\sec t$ and $\tan t$ can be rewritten following the identities: $$\sec t=\frac{1}{\cos t}\hspace{2cm}\tan t=\frac{\sin t}{\cos t}$$ So, $$X=\frac{1}{\frac{1}{\cos t}+\frac{\sin t}{\cos t}}$$ $$X=\frac{1}{\frac{1+\sin t}{\cos t}}$$ $$X=\frac{\cos t}{1+\sin t}$$ Now multiply both the numerator and denominator by $(1-\sin t)$: $$X=\frac{\cos t(1-\sin t)}{(1+\sin t)(1-\sin t)}$$ $$X=\frac{\cos t(1-\sin t)}{1-\sin^2 t}$$ (do not forget that $(A-B)(A+B)=A^2-B^2$) $$X=\frac{\cos t(1-\sin t)}{\cos^2t}$$(Pythagorean identity: $1-\sin^2 t=\cos^2t$) $$X=\frac{1-\sin t}{\cos t}$$ Therefore the left and right sides are equal to each other. $$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$ The equation is an identity.
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