Answer
$\dfrac{\sin2x}{\sin x}-\dfrac{\cos 2x}{\cos x}=\sec x$
Work Step by Step
$\dfrac{\sin2x}{\sin x}-\dfrac{\cos 2x}{\cos x}=\sec x$
Substitute $\sin2x$ by $2\sin x\cos x$ and $\cos2x$ by $1-2\sin^{2}x$ (These are the double angle formulas for sine and cosine):
$\dfrac{2\sin x\cos x}{\sin x}-\dfrac{1-2\sin^{2}x}{\cos x}=\sec x$
$2\cos x-\dfrac{1-2\sin^{2}x}{\cos x}=\sec x$
Evaluate the subtraction on the left:
$\dfrac{2\cos^{2}x-1+2\sin^{2}x}{\cos x}=\sec x$
Factor the numerator by just grouping the sine and cosine terms:
$\dfrac{2(\sin^{2}x+\cos^{2}x)-1}{\cos x}=\sec x$
Substitute $\sin^{2}x+\cos^{2}x$ by $1$ and simplify:
$\dfrac{2(1)-1}{\cos x}=\sec x$
$\dfrac{1}{\cos x}=\sec x$
Since $\dfrac{1}{\cos x}=\sec x$, the identity is proved.
$\sec x=\sec x$
