(c) Since, $\sin{\theta = \sin\alpha}$ …(i) And $\cos{\theta = \cos\alpha}$ …(ii) [divided Eq. (i) by Eq. (ii)] $\therefore\ \ \tan{\theta = \tan\alpha}$ $\Rightarrow \ \ \theta = n\pi + \alpha$
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Ph.D. in Physics with 19+ years of teaching experience.