## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 228: 69

#### Answer

$$\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}=\frac{\sin s}{\sin t\cos t}$$ The equation is proved to be an identity.

#### Work Step by Step

$$\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}=\frac{\sin s}{\sin t\cos t}$$ The left side is more complicated, so it should be dealt with first $$\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}$$ $$=\frac{\sin s\cos t-\sin t\cos s}{\sin t}+\frac{\cos s\cos t+\sin s\sin t}{\cos t}$$ $$=\frac{(\sin s\cos t-\sin t\cos s)\cos t+(\cos s\cos t+\sin s\sin t)\sin t}{\sin t\cos t}$$ $$=\frac{\sin s\cos^2 t-\sin t\cos t\cos s+\sin t\cos t\cos s+\sin s\sin^2 t}{\sin t\cos t}$$ $$=\frac{\sin s\cos^2 t+\sin s\sin^2 t}{\sin t\cos t}$$ (2 products in the middle are eliminated) $$=\frac{\sin s(\cos^2 t+\sin^2 t)}{\sin t\cos t}$$ $$=\frac{\sin s\times 1}{\sin t\cos t}$$ (for $\sin^2 t+\cos^2 t=1$) $$=\frac{\sin s}{\sin t\cos t}$$ Therefore, the equation is indeed an identity.

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