Question

If $5\tan \theta =4$, then $\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }=$

• A. $0$
• B. $1$
• C. $\frac{1}{6}$
• D. $6$

Answer 1

Answer: (C)

$5\tan \theta =4\Rightarrow \tan \theta =\frac{4}{5}$
$\therefore \sin \theta =\frac{4}{\sqrt{41}}$ and $\cos \theta =\frac{5}{\sqrt{41}}$
$\frac{5\sin \theta -3\cos \theta }{5\sin 0+2\cos 0}=\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{5}{\sqrt{41}}}{5\times \frac{4}{\sqrt{41}}+2\times \frac{5}{\sqrt{41}}}=\frac{1}{6}.$