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!
[Math] Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$
proof-writingtrigonometry
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Best Answer
Your solution looks correct. An alternative would be to multiply numerator and denominator by $\cot \theta$ to get $$\frac{1}{1 + cot^2 \theta}$$
and then multiplying by $\sin^2 \theta$ we have $$\frac{1}{1 + cot^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta + \cos^2 \theta} = \sin^2 \theta$$