#### Answer

$1-\tan ^{2}\dfrac {\theta }{2}=\dfrac {2\cos \theta }{1+\cos \theta }$

#### Work Step by Step

$1-\tan ^{2}\dfrac {\theta }{2}=1-\dfrac {\sin ^{2}\dfrac {\theta }{2}}{\cos ^{2}\dfrac {\theta }{2}}=\dfrac {\cos ^{2}\dfrac {\theta }{2}-\sin ^{2}\dfrac {\theta }{2}}{\cos ^{2}\dfrac {\theta }{2}}=\dfrac {\cos \left( 2\times \dfrac {\theta }{2}\right) }{\dfrac {1+\cos \theta }{2}}=\dfrac {2\cos \theta }{1+\cos \theta }$