Question

If tan $A=\cot B$ where $A$ and $B$ are acute angles, prove that $A+B={90}^{\circ }.$

Answer 1

$tanA=cotBtanAtanB=1thentan(A+B)=\left(\frac{tanA+tanB}{1-tanAtanB}\right)\left(\frac{1}{tan(a+B)}\right)=\left(\frac{0}{tanA+tanB}\right)\therefore tan(A+B)=\infty =undefindthenA+B=\left(\frac{\pi }{2}\right)$