prove that tan(90-beta)=cot beta
Follow Math Help Forum on Facebook and Google+
$\displaystyle \tan(90-\beta)=\frac{\sin(90-\beta}{\cos(90-\beta)}=\frac{\sin 90\cos\beta-\sin\beta\cos 90}{\cos 90\cos\beta+\sin 90\sin\beta}=\frac{\cos\beta}{\sin\beta}=\cot\beta$
View Tag Cloud