Prove that `sintheta+sin3theta+sin5theta+.....+sin(2n-1)theta=(sin^2ntheta)/(sintheta)dot`

Explanation

To prove that

\[ \sin \theta + \sin 3\theta + \sin 5\theta + \ldots + \sin(2n-1)\theta = \frac{\sin^2 n\theta}{\sin \theta} \]

we can use the formula for the sum of sines.

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