Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Summary Exercises on Verifying Trigonometric Identities - Page 239: 18

Answer

$$\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}=2\csc s$$ As 2 sides are proved equal below, the equation is an identity.

Work Step by Step

$$\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}=2\csc s$$ The left side, being more complex, would be dealt with first. $$X=\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}$$ $$X=\frac{\sin^2s+(1+\cos s)^2}{\sin s(1+\cos s)}$$ Recall that $(A+B)^2=A^2+2AB+B^2$, so we can expand $(1+\cos s)^2$ as follows. $$X=\frac{\sin^2s+1+2\cos s+\cos^2s}{\sin s(1+\cos s)}$$ $$X=\frac{(\sin^2s+\cos^2s)+1+2\cos s}{\sin s(1+\cos s)}$$ For $(\sin^2s+\cos^2s)$, it equals $1$, said by the Pythagorean Identities. $$X=\frac{1+1+2\cos s}{\sin s(1+\cos s)}$$ $$X=\frac{2+2\cos s}{\sin s(1+\cos s)}$$ $$X=\frac{2(1+\cos s)}{\sin s(1+\cos s)}$$ $$X=\frac{2}{\sin s}$$ Recall that $\frac{1}{\sin s}=\csc s$ (Reciprocal Identity) $$X=2\csc s$$ So, $$\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}=2\csc s$$ The equation is an identity, since 2 sides are proved equal.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.