#### Answer

$\tan \left( \dfrac {x}{2}+\dfrac {\pi }{4}\right) =\dfrac {1+2\sin \dfrac {x}{2}\cos \dfrac {x}{2}}{\cos x}=secx+\tan x $

#### Work Step by Step

$\tan \left( \dfrac {x}{2}+\dfrac {\pi }{4}\right) =\dfrac {\tan \dfrac {x}{2}+\tan \dfrac {\pi }{4}}{1-\tan \dfrac {x}{2}\tan \dfrac {\pi }{4}}=\dfrac {1+\tan \dfrac {x}{2}}{1-\tan \dfrac {x}{2}}=\dfrac {\left( 1+\tan \dfrac {x}{2}\right) \left( 1+\tan \dfrac {x}{2}\right) }{\left( 1-\tan \dfrac {x}{2}\right) \left( 1+\tan \dfrac {x}{2}\right) }=\dfrac {1+\tan ^{2}\dfrac {x}{2}+2\tan \dfrac {x}{2}}{1-\tan ^{2}\dfrac {x}{2}}=\dfrac {\dfrac {\dfrac {1}{\cos ^{2}\dfrac {x}{2}}+2\tan \dfrac {x}{2}}{\cos ^{2}\dfrac {x}{2}-\sin ^{2}\dfrac {x}{2}}}{\cos ^{2}\dfrac {x}{2}}=\dfrac {1+2\sin \dfrac {x}{2}\cos \dfrac {x}{2}}{\cos x}=secx+\tan x $