Question

If $3\cot \theta =2$, find the value of $\frac{2\sin \theta -3\cos \theta }{2\sin \theta +3\cos \theta }$

Answer 1

$\cot =\frac{2}{3}=\frac{2\sin \theta -3\cos \theta }{2\sin \theta +3\cos \theta }$

$=\frac{2-3\cot \theta }{2+3\cos \theta }=\frac{2-2}{2+2}=0$.