## Precalculus (6th Edition)

$\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}=\frac{\sin s}{\sin t\cos t}$
Start with the left side: $\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}$ Expand using the sum and difference formulas for sine and cosine: $=\frac{\sin s\cos t-\cos s\sin t}{\sin t}+\frac{\cos s\cos t+\sin s\sin t}{\cos t}$ Get a common denominator: $=\frac{(\sin s\cos t-\cos s\sin t)\cos t}{\sin t\cos t}+\frac{(\cos s\cos t+\sin s\sin t)\sin t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t}{\sin t\cos t}+\frac{\cos s\cos t\sin t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t+\cos s\cos t\sin t+\sin s\sin^2 t}{\sin t\cos t}$ Note that the second and third terms in the numerator cancel. Simplify: $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t+\cos s\sin t\cos t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s(\cos^2+\sin^2 t)}{\sin t\cos t}$ $=\frac{\sin s*1}{\sin t\cos t}$ $=\frac{\sin s}{\sin t\cos t}$ Since this equals the right side, the identity has been proven.