Question

If 3 $\cot \theta$ = 2, find the value of $\frac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }$

Answer 1

Given,

$3\cot \theta =2$

$\cot \theta =\frac{2}{3}$

$\tan \theta =\frac{3}{2}$

$\frac{4sin\theta -3cos\theta }{2sin\theta +6cos\theta }$

$=\frac{4sin\theta -3cos\theta }{2sin\theta +6cos\theta }\times \frac{\sin \theta }{\sin \theta }$ ---------- ( Multiply and divide by $\sin \theta$ )

$=\frac{\frac{4sin\theta -3cos\theta }{sin\theta }}{\frac{2sin\theta +6cos\theta }{sin\theta }}$

$=\frac{4-3\left(\frac{cos\theta }{sin\theta }\right)}{2+6\left(\frac{cos\theta }{sin\theta }\right)}$

$=\frac{4-3cot\theta }{2+6cot\theta }$

$=\frac{4-3\times \frac{2}{3}}{2+6\times \frac{2}{3}}$

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

$=\frac{2}{6}$

$=\frac{1}{3}$