Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:

Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$

My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!

"I know! Let's stand him on his head!"
$$ \frac{\tan \theta \ + \ \cot \theta}{\sec \theta \ \cdot \ \sin \theta} \ = \ \frac{\tan \theta \ + \ \cot \theta}{\tan \theta } \ = \ 1 \ + \ \frac{\cos^2 \theta }{\sin^2 \theta } \ = \ \frac{\sin^2 \theta \ + \ \cos^2 \theta }{\sin^2 \theta } \ = \ \frac{1}{\sin^2 \theta } $$
$$ \Rightarrow \ \ \frac{\sec \theta \ \cdot \ \sin \theta}{\tan \theta \ + \ \cot \theta} \ = \ \sin^2 \theta \ \ . $$
As I look through the thread again, this is rather like dani_s's proof, without involving cotangent...
