$$\sum_{k=1}^n \frac{1}{\sin k\theta \sin (k+1)\theta} = \dfrac{1}{\sin\theta\cdot \sin2\theta} + \dfrac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$$ up to $n$ terms.
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1"I tried but in vain". In order to help you (and not to do your homework), it is important you tell what you tried and where you are stuck. – Claude Leibovici May 29 '14 at 08:20
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Related : http://math.stackexchange.com/questions/464031/find-the-sum-frac1-cos0-circ-cos1-circ-frac1-cos1-circ-cos2-cir – lab bhattacharjee May 29 '14 at 09:45
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Note $$\dfrac{\sin{\theta}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\dfrac{\sin{((k+1)\theta-k\theta)}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\cot{(k\theta)}-\cot{(k+1)\theta}$$
math110
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1New users who have no post history and do not show evidence of what they tried, should not receive free answers. – heropup May 29 '14 at 07:55
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Here is a hint: the sum telescopes, with some careful manipulation:
$$\sin \theta = \sin((k+1)\theta - k\theta) = \sin(k+1)\theta \cos k\theta - \sin k\theta \cos (k+1)\theta.$$
heropup
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It differs to the extent that I don't explicitly write out the form of the telescoping terms. I think that @math110's answer gives away too much, whereas some additional thought is required with my hint. – heropup May 29 '14 at 08:18
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