Question

If $\tan \theta =\frac{4}{3}$, then find the value of $\frac{4\sin \theta +3\cos \theta }{4\sin \theta -3\cos \theta }$ ?

• A. $0$
• B. $1$
• C. $\frac{16}{9}$
• D. $\frac{25}{7}$

Answer 1

Answer: (D)

Given : $\tan \theta =\frac{4}{3}$

Consider, $\frac{4\sin \theta +3\cos \theta }{4\sin \theta -3\cos \theta }$

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

$=\frac{4\tan \theta +3}{4\tan \theta -3}$

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

$=\frac{16+9}{16-9}=\frac{25}{7}$

Hence, $\frac{4\sin \theta +3\cos \theta }{4\sin \theta -3\cos \theta }=\frac{25}{7}$.