## Algebra 2 (1st Edition)

$-\sqrt 2-1$
Double -angle Theorem can be defined as: $\tan \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{1+ \cos \theta}}$ $\tan 112.5 ^{\circ}=\tan \dfrac{225^{\circ}}{2}$ Since $\tan$ is Negative in the second quadrant. Thus, $- \sqrt {\dfrac{1- \cos 225 ^{\circ}}{1+ \cos 225^{\circ}}}= \sqrt {\dfrac{1- \cos (180 ^{\circ}+45 ^{\circ})}{1+ \cos (180 ^{\circ}+45 ^{\circ})}}=- \sqrt {\dfrac{1+ \cos 45 ^{\circ}}{1- \cos 45 ^{\circ}}}$ and $-\sqrt {\dfrac{1+ (1/\sqrt 2)}{1- (1/\sqrt 2)}}=-\sqrt {\dfrac{\sqrt 2+1}{\sqrt 2-1}}$ Hence, $\tan 112.5 ^{\circ}=-\sqrt 2-1$