A) \[\frac{3\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: C
Solution :
Given, \[y=3\sin \theta .\cos \theta \] \[\frac{dy}{d\theta }=3[\sin \theta (-\sin \theta )+\cos \theta (cos\theta )]\] \[\frac{dy}{d\theta }=3[co{{s}^{2}}\theta -{{\sin }^{2}}\theta ]=3cos2\theta \]and \[x={{e}^{\theta }}\sin \theta \] \[\frac{dx}{d\theta }={{e}^{\theta }}\cos \theta +\sin \theta \,{{e}^{\theta }}\] \[\frac{dx}{d\theta }={{e}^{\theta }}(\sin \theta +\cos \theta )\] ?.(2) Dividing (1) by (2) \[\frac{dy}{dx}=\frac{3\cos 2\theta }{{{e}^{\theta }}(sin\theta +cos\theta )}=\frac{3({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )}{{{e}^{\theta }}(sin\theta +cos\theta )}\] \[\frac{dy}{dx}=\frac{3(\cancel{\cos \theta +\sin \theta })(cos\theta -sin\theta )}{{{e}^{\theta }}(\cancel{sin\theta +cos\theta })}\] \[\frac{dy}{dx}=\frac{3(\cos \theta -\sin \theta )}{{{e}^{\theta }}}\] Given tangent is parallel to x-axis then\[\frac{dy}{dx}=0\] \[0=\frac{3(cos\theta -sin\theta )}{{{e}^{\theta }}}\] or\[\cos \theta =\sin \theta =0\Rightarrow \cos \theta =\sin \theta \] \[\Rightarrow \tan \theta =1\Rightarrow \tan \theta =\frac{\tan -}{4}\Rightarrow \theta =\frac{\pi }{4}\]
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