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dalendrk
[tex]We\ know:\\\\\sin\left(\dfrac{\pi}{2}-x\right)=\cos x\\and\\\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\---------------------------\\\\\dfrac{\sin^2\left(\dfrac{\pi}{2}-x\right)}{\sin x}=\dfrac{\cos^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}\\\\=\dfrac{1}{\sin x}-\sin x=\csc x-\sin x[/tex]

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