Question

Find the acute angles A and B satisfying $\sec A\cdot \cot B-\sec A-2\cdot \cot B+2=0$.

Answer 1

$\frac{\cos B}{\sin B.\cos B}-\frac{1}{\cos A}-\frac{2\cos B}{\sin B}+2=\frac{1}{\cos A}[\frac{\cos B}{\sin B}-1]-2[\frac{\cos B}{\sin B}-1]=[\cot B-1][\sec A-2]=0=\cot B=1;B=\frac{\pi }{4}\sec A=2;A=\frac{\pi }{3}$