## Precalculus (6th Edition)

$\frac{\cos\theta+1}{\tan^2\theta}=\frac{\cos\theta}{\sec\theta-1}$
Simplify the left side: $\frac{\cos\theta+1}{\tan^2\theta}$ Rewrite it in terms of sine and cosine: $\frac{\cos\theta+1}{\frac{\sin^2\theta}{\cos^2\theta}}$ Multiply top and bottom by $\cos^2\theta$: $\frac{\cos\theta+1}{\frac{\sin^2\theta}{\cos^2\theta}}*\frac{\cos^2\theta}{\cos^2\theta}$ $=\frac{(\cos\theta+1)\cos^2\theta}{\sin^2\theta}$ Rewrite $\sin^2\theta$ as $1-\cos^2\theta$: $=\frac{(\cos\theta+1)\cos^2\theta}{1-\cos^2\theta}$ Factor the denominator and simplify: $=\frac{(\cos\theta+1)\cos^2\theta}{(1+\cos\theta)(1-\cos\theta)}$ $=\frac{(\cos\theta+1)\cos^2\theta}{(\cos\theta+1)(1-\cos\theta)}$ $=\frac{\cos^2\theta}{1-\cos\theta}$ Simplify the right side: $\frac{\cos\theta}{\sec\theta-1}$ Rewrite it in terms of sine and cosine: $=\frac{\cos\theta}{\frac{1}{\cos\theta}-1}$ Multiply top and bottom by $\cos\theta$: $=\frac{\cos\theta}{\frac{1}{\cos\theta}-1}*\frac{\cos\theta}{\cos\theta}$ $=\frac{\cos^2\theta}{1-\cos\theta}$ Since the left side and the right side both simplify to $=\frac{\cos^2\theta}{1-\cos\theta}$, they are equal to each other, and the identity is proven.