If \[2\cot \theta =3,\]then what is \[\frac{2\cos \theta -\sin \theta }{2\cos \theta +\sin \theta }\]equal to? |
A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
In \[\Delta ABC,\]\[\cot =\frac{3}{2}=\frac{AB}{AC}\] |
\[\therefore \]\[AB=3\]and \[AC=2\] |
By Pythagoras theorem, |
\[B{{C}^{2}}={{(2)}^{2}}+{{(3)}^{2}}\] |
\[\Rightarrow \] \[BC=\sqrt{13}\] |
Now, \[\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}=\frac{AB}{AC}=\frac{3}{\sqrt{13}}\] |
and \[\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{AC}{BC}=\frac{2}{\sqrt{13}}\] |
\[\therefore \]\[\frac{2\cos \theta -\sin \theta }{2\cos \theta +\sin \theta }=\frac{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}=\frac{4}{8}=\frac{1}{2}\] |
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