Sum of Squared Sines

Using the formula for an infinite geometric series ,

$\begin{aligned} \sum_{n = 1}^{\infty} \sin^{2n}\theta &= \sin^2\theta + \sin^4\theta + \sin^6\theta + \ldots \\ \\ &= \frac{\sin^2\theta}{1 - \sin^2\theta} \\ \\ &= \frac{\sin^2\theta}{\cos^2 \theta} = \boxed{\tan^2\theta} \end{aligned}$

I realize that this solution abuses the fact that we are given four answer choices from which to select, but it seems a valid strategy for the problem as posed.

Observe that $\theta=0$ has $\lvert\sin\theta\rvert\neq 1$ , so we may consider the sum at $\theta=0$ . When $\theta=0$ , $\sin\theta=0$ , so all powers of $\sin\theta$ must also equal zero. Therefore the sum must equal 0, yet of the answer choices, only $\tan^2\theta=0$ when $\theta=0$ , thus $\boxed{\tan^2\theta}$ must be the answer.

BY the formula for sum of infinite gp. We have sum, S = a/(1-r), where a is the first term and r is the common ratio. Now here in the ques., a=(sin theta)^2= r , Therefore S= (sin theta)^2/1-(sin theta)^2 =(tan theta)^2.ans.