#### Answer

$\tan \dfrac {147}{2}=\tan 73.5$

#### Work Step by Step

$\sqrt {\dfrac {1-\cos 147}{1+\cos 147}}==\sqrt {\dfrac {1-\cos \left( 2\times \dfrac {147}{2}\right) }{1+\cos \left( 2\times \dfrac {147}{2}\right) }}=\sqrt {\dfrac {1-\left( \cos ^{2}\dfrac {147}{2}-\sin ^{2}\dfrac {147}{2}\right) }{1+\left( \cos ^{2}\dfrac {147}{2}-\sin ^{2}\dfrac {147}{2}\right) }}=\sqrt {\dfrac {2\sin ^{2}\dfrac {147}{2}}{2\cos ^{2}\dfrac {147}{2}}}=\tan \dfrac {147}{2}=\tan 73.5$