Answer
See the steps.
Work Step by Step
$LHS =\dfrac{1-\tan{x}}{1+\tan{x}} =\dfrac{1-\dfrac{\sin{x}}{\cos{x}}}{1+\dfrac{\sin{x}}{\cos{x}}}$
$LHS = \dfrac{\cos{x} -\sin{x}}{\cos{x} +\sin{x}}$
Multiply by $\dfrac{\cos{x} -\sin{x}}{\cos{x} -\sin{x}}$
$LHS =\dfrac{\cos^2{x}-2\sin{x}\cos{x}+\sin^2{x}}{\cos^2{x}-\sin^2{x}}$
$LHS =\dfrac{1-\sin{2x}}{\cos{2x}}= RHS$