Answer
$\frac{\cos\theta+1}{\tan^2\theta}=\frac{\cos\theta}{\sec\theta-1}$
Work Step by Step
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.
