Answer:
\[\sqrt{3}\text{ }sin\text{ }\theta =cos\text{ }\theta \] [Given] \[\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}\] or \[\tan \,\theta =\frac{1}{\sqrt{3}}\] \[\Rightarrow \tan \theta =\tan 30{}^\circ \Rightarrow \theta =30{}^\circ \] Now, \[\frac{3\,{{\cos }^{2}}\theta +2\cos \theta }{3\cos \theta +2}=\frac{\cos \theta (3cos\theta +2)}{(3cos\theta +2)}\] \[=\cos \theta \] Put \[\theta =30{}^\circ \] \[\Rightarrow \cos 30{}^\circ =\frac{\sqrt{3}}{2}\]
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