Question

If $3\cot \theta -4=0$, then the value of $\frac{3\cot \theta +4\cos \theta }{3\sin \theta -2\cos }$ will be

• A. $\frac{4}{3}$
• B. $25$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$25$

$3\cot \theta -4=0$
$\cot \theta =\frac{4}{3}$
Now,$\frac{3\sin \theta +4\cos \theta }{3\sin \theta -2\cos \theta }$
$=\frac{\sin \theta [3+4\frac{\cos \theta }{\sin \theta }]}{\sin \theta [3-2\frac{\cos \theta }{\sin \theta }]}$
$=\frac{3+4\cot \theta }{3-2\cot \theta }$
$=\frac{3+4\times \frac{4}{3}}{3-2\times \frac{4}{3}}$
$=\frac{3+\frac{16}{3}}{3-\frac{8}{3}}=\frac{\frac{25}{3}}{\frac{1}{3}}$
Hence, $=\frac{3\sin \theta +4\cos \theta }{3\sin \theta -2\cos \theta }=25$