A) \[\frac{1}{3}\]
B) \[-3\]
C) 3
D) 2
Correct Answer: C
Solution :
| \[\sec \theta =\frac{13}{5}\] |
| \[\therefore \]\[\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}=\sqrt{\frac{169}{25}-1}=\frac{12}{5}\] |
| \[\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }=\frac{2\frac{\sin \theta }{\cos \theta }-3}{4\frac{\sin \theta }{\cos \theta }-9}\] |
| Divide numerator and denominator by \[\cos \theta ,\] |
| \[=\frac{2\tan \theta -3}{4\tan \theta -9}=\frac{2\times \frac{12}{5}-3}{4\times \frac{12}{5}-9}=\frac{9/5}{3/5}=\frac{9}{3}=3\] |
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