question_answer

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]

\[=\frac{20-15}{20+10}=\frac{5}{30}=\frac{1}{6}\]