If \[5\tan \theta =4,\] then \[\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }\] is equal to |
A) \[\frac{1}{4}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
Given, \[5\tan \theta =4\]\[\Rightarrow \]\[\frac{5\sin \theta }{\cos \theta }=4\] |
\[\Rightarrow \] \[5\sin \theta =4cos\theta \] ... (i) |
Now, \[\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }=\frac{4\cos \theta -3\cos \theta }{4\cos \theta +2\cos \theta }\] |
[from Eq. (i)] |
\[=\frac{\cos \theta }{6\cos \theta }=\frac{1}{6}\] |
Alternate Method |
Given, \[5\tan \theta =4\] |
\[\Rightarrow \] \[\tan \theta =\frac{4}{5}\] |
\[\because \] \[\tan \theta =\frac{BC}{AB}\] |
Similarly, |
\[\therefore \] \[\sin \theta =\frac{4}{\sqrt{41}}\]\[\Rightarrow \]\[\cos \theta =\frac{5}{\sqrt{41}}\] |
\[\therefore \] \[\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }\,\,=\,\,\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{5}{\sqrt{41}}}{5\times \frac{4}{\sqrt{41}}+2\times \frac{5}{\sqrt{41}}}\] |
[putting values] |
\[=\frac{20-15}{20+10}=\frac{5}{30}=\frac{1}{6}\] |
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