# Thread: Prove: (cosxcotx)/1-sinx - 1 = csc x

1. ## Prove: (cosxcotx)/1-sinx - 1 = csc x

As the title says, Prove: (cosxcotx)/1-sinx - 1 = csc x

Any help is appreciated! Thanks.

2. Hello, SMA777!

This is a strange one . . .

Prove: . $\frac{\cos x\cot x}{1-\sin x} - 1 \:= \:\csc x$

On the left side, multiply the fraction by $\frac{1+\sin x}{1+\sin x}$

. $\frac{\cos x\!\cdot\!\dfrac{\cos x}{\sin x}}{1-\sin x}\cdot{\color{blue}\frac{1+\sin x}{1+\sin x}} -1 \;=\;\frac{\dfrac{\cos^2\!x}{\sin x}(1 + \sin x)}{1 - \sin^2\!x}-1 \;=\;\frac{\dfrac{\cos^2\!x}{\sin x}(1 + \sin x)}{\cos^2x} -1$

. . $= \;\frac{1 + \sin x}{\sin x} - 1 \;=\;\frac{1+\sin x - \sin x}{\sin x} \;=\; \frac{1}{\sin x} \;=\;\csc x$

3. Ah, thank you!

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# (sinx-1)(cosecx 1)=cosxcotx

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