Answer
$\dfrac {2}{1+\cos x}-\tan ^{2}\dfrac {x}{2}=1$
Work Step by Step
$\dfrac {2}{1+\cos x}-\tan ^{2}\dfrac {x}{2}=\dfrac {2}{1+\cos x}-\dfrac {\sin ^{2}\dfrac {x}{2}}{\cos ^{2}\dfrac {x}{2}}=\dfrac {2}{1+\cos x}-\dfrac {\left( \sqrt {\dfrac {1-\cos x}{2}}\right) ^{2}}{\left( \sqrt {\dfrac {1+\cos x}{2}}\right) ^{2}}=\dfrac {2}{1+\cos x}-\dfrac {1-\cos x}{1+\cos x}=\dfrac {1+\cos x}{1+\cos x}=1$
