Question

Given that $4\cot =3,Evaluate\frac{2\sin A+3\cos A}{4\sin A-5\cos A}$

Answer 1

Take $\sin A$ common from numerator and denominator of the given expression,

$\frac{2\sin A+3\cos A}{4\sin A-5\cos A}$

$=\frac{\sin A(2+3\frac{\cos A}{\sin A})}{\sin A(4-5\frac{\cos A}{\sin A})}$

$=\frac{2+3\cot A}{4-5\cot A}$

$=\frac{2+3\times \frac{3}{4}}{4-5\times \frac{3}{4}}$

$=\frac{2+\frac{9}{4}}{4-\frac{15}{4}}$

$=\frac{17}{4}\times \frac{4}{1}=17$