Answer
$\dfrac{1+\sec x}{\sec x}=\dfrac{\sin^{2}x}{1-\cos x}$
Work Step by Step
$\dfrac{1+\sec x}{\sec x}=\dfrac{\sin^{2}x}{1-\cos x}$
Substitute $\sin^{2}x$ by $1-\cos^{2}x$:
$\dfrac{1+\sec x}{\sec x}=\dfrac{1-\cos^{2}x}{1-\cos x}$
Factor the numerator on the right side:
$\dfrac{1+\sec x}{\sec x}=\dfrac{(1-\cos x)(1+\cos x)}{1-\cos x}$
Simplify the right side:
$\dfrac{1+\sec x}{\sec x}=1+\cos x$
Rewrite the left side as $\dfrac{1}{\sec x}+\dfrac{\sec x}{\sec x}$ and simplify it:
$\dfrac{1}{\sec x}+1=1+\cos x$
Substitute $\sec x$ by $\dfrac{1}{\cos x}$:
$1+\dfrac{1}{\Big(\dfrac{1}{\cos x}\Big)}=1+\cos x$
Evaluate the division on the left side and the identity will be proved:
$1+\cos x=1+\cos x$
