question_answer

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       \[\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{AC}{BC}=\frac{2}{\sqrt{13}}\]

\[\therefore \]\[\frac{2\cos \theta -\sin \theta }{2\cos \theta +\sin \theta }=\frac{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}=\frac{4}{8}=\frac{1}{2}\]