A) 0
B) 1
C) 1/6
D) 6
Correct Answer: C
Solution :
\[5\tan \theta =4\Rightarrow \tan \theta =\frac{4}{5}\] \[\therefore \sin \theta =\frac{4}{\sqrt{41}}\]and \[\cos \theta =\frac{5}{\sqrt{41}}\] \[\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }=\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{5}{\sqrt{41}}}{5\times \frac{4}{\sqrt{41}}+2\times \frac{5}{\sqrt{41}}}\] \[\frac{20-15}{20+10}=\frac{5}{30}=\frac{1}{6}\].
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