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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!

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  • $\begingroup$ Quite simple question, jus convert everything into sin and cos forms. It will be easy to solve $\endgroup$ – Kailas Mar 12 '14 at 6:10
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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 $$\frac{1}{1 + cot^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta + \cos^2 \theta} = \sin^2 \theta$$

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    $\begingroup$ Very clever, I must say. Definitely much shorter than my proof $\endgroup$ – TrueDefault Mar 12 '14 at 6:30
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Proof: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\dfrac{\left(\dfrac{1}{\cos\theta}\cdot\sin\theta\right)}{\tan\theta+\cot\theta}$$ $$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}\right)}$$ $$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin^2\theta}{\sin\theta\cdot\cos\theta}+\dfrac{\cos^2\theta}{\sin\theta\cdot\cos\theta}\right)}$$ $$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cdot\cos\theta}\right)}$$ $$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{1}{\sin\theta\cdot\cos\theta}\right)}$$ $$=\dfrac{\sin\theta}{\cos\theta}\cdot\left(\sin\theta\cdot\cos\theta\right)$$ $$=\dfrac{\sin\theta}{\cos\theta}\cdot\sin\theta\cdot\cos\theta$$ $$=\sin^2\theta$$ $$\displaystyle \boxed{\therefore \dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta}$$

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$\dfrac{\sec \theta \cdot \sin\theta}{\tan\theta+\cot\theta}=\dfrac{\sec \theta \cdot \sin\theta}{\tan\theta+\frac{1}{\tan\theta}}$ $= \dfrac{\sec \theta \cdot \sin\theta\cdot \tan\theta}{\tan^2\theta+1}$ $=\dfrac{\sec\theta\cdot\sin\theta\cdot\frac{\sin\theta}{\cos\theta}}{\sec^2\theta}$ $=\dfrac{\sin\theta\cdot\sin\theta}{\sec\theta\cdot\cos\theta}$ $=\dfrac{\sin^2\theta}{1}=\sin^2\theta$.

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"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...

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 secθ⋅sinθ/(tanθ+cotθ)
    =(1/cosθ⋅sinθ)/(tanθ+1/tanθ)
    =(sinθ/cosθ)/(((tanθ)^2+1)/tanθ)
    =(tanθ)/(((secθ)^2)/tanθ)
    =(tanθ)^2/(secθ)^2
    =((sinθ)^2/(cosθ)^2)/(1/(cosθ)^2)
    =(sinθ)^2
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