Question

If $7\tan \theta =4$, then $\frac{(7\sin \theta -3\cos \theta )}{(7\sin \theta +3\cos \theta )}=?$
(a) $\frac{1}{7}$
(b) $\frac{5}{7}$
(c) $\frac{3}{7}$
(d) $\frac{5}{14}$

Answer 1

Correct option is (a) $\frac{1}{7}$

Given: $7\tan \theta =4$

$\Rightarrow \tan \theta =\frac{4}{7}$

Consider, $\frac{(7\sin \theta -3\cos \theta )}{(7\sin \theta +3\cos \theta )}$

Dividing numerator and denominator by $\cos \theta$

$=\frac{\frac{7\sin \theta -3\cos \theta }{\cos \theta }}{\frac{7\sin \theta +3\cos \theta }{\cos \theta }}$

$=\frac{7\tan \theta -3}{7\tan \theta +3}$

$=\frac{7\left(\frac{4}{7}\right)-3}{7\left(\frac{4}{7}\right)+3}$

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

$=\frac{1}{7}$

$\therefore \frac{(7\sin \theta -3\cos \theta )}{(7\sin \theta +3\cos \theta )}=\frac{1}{7}$