A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
| Given, \[\sec \theta =\frac{13}{15}\]\[\Rightarrow \]\[{{\sec }^{2}}\theta =\frac{169}{25}\] |
| \[\Rightarrow \]\[1+{{\tan }^{2}}\theta =\frac{169}{25}\]\[\Rightarrow \]\[{{\tan }^{2}}\theta =\frac{169}{25}-1\] |
| \[\Rightarrow \]\[{{\tan }^{2}}\theta =\frac{144}{25}\] |
| \[\Rightarrow \] \[\tan \theta =\frac{12}{5}\] ?(i) |
| \[\therefore \] \[\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }=\frac{2\cdot \frac{\sin \theta }{\cos \theta }-3}{4\cdot \frac{\sin \theta }{\cos \theta }-9}\] |
| \[=\frac{2\tan \theta -3}{4\tan \theta -9}=\frac{2\left( \frac{12}{5} \right)-3}{4\left( \frac{12}{5} \right)-9}\] [from Eq. (i)] |
| \[=\frac{24-15}{48-45}=\frac{9}{3}=3\] |
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