Question

If $3\cot \theta =4$, find the value of $\frac{5\sin \theta -3\cos \theta }{5\sin \theta +3\cos \theta }$

• A. $\frac{1}{9}$
• B. $\frac{2}{7}$
• C. $3$
• D. $4$

Answer 1

The correct option is A $\frac{1}{9}$

Given, $3\cot \theta =4$

$\therefore \cot \theta =\frac{4}{3}$

Now, $\frac{5\sin \theta -3\cos \theta }{5\sin \theta +3\cos \theta }$

Dividing both numerator and denominator by $\sin \theta$, we get

$=\frac{5\frac{\sin \theta }{\sin \theta }-\frac{3\cos \theta }{\sin \theta }}{5\frac{\sin \theta }{\sin \theta }+\frac{3\cos \theta }{\sin \theta }}$

$=\frac{5-3\cot \theta }{5+3\cot \theta }$

$=\frac{5-4}{5+4}$ .......$[\because 3\cot \theta =4]$

$=\frac{1}{9}$

$\therefore \frac{5\sin \theta -3\cos \theta }{5\sin \theta +3\cos \theta }=\frac{1}{9}$