Aus Online Mathematik Brückenkurs 1

By subtracting \displaystyle 360^{\circ } from \displaystyle \text{495}^{\circ }, we do not change the value of the tangent:

\displaystyle \tan \text{495}^{\circ }=\tan \left( \text{495}^{\circ }-360^{\circ } \right)=\tan \text{135}^{\circ }

We know from exercise a that \displaystyle \cos 135^{\circ }=-\frac{1}{\sqrt{2}} and \displaystyle \sin 135^{\circ }=\frac{1}{\sqrt{2}}, which gives

\displaystyle \tan 135^{\circ }=\frac{\sin 135^{\circ }}{\cos 135^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1