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I have the trigonometric expression: $$\csc x+\cot x$$ And I have to manipulate it into: $$\frac1{\csc x-\cot x}$$ I've tried, of course, changing the $\csc$ and $\cot$ into $\frac1\sin$ and $\frac1\tan$, but it's gotten me nowhere. I suspect I have to multiply it by some expression over itself but I'm not sure what or how.

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    $\begingroup$ $$\csc^2x-\cot^2x=?$$ $\endgroup$ – lab bhattacharjee Feb 13 '18 at 12:56
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Notice that the difference of two squares: $$(\csc x+\cot x)(\csc x-\cot x)=\csc^2x-\cot^2x=\frac1{\sin^2x}-\frac{\cos^2x}{\sin^2x}=\frac{\sin^2x}{\sin^2x}=1$$ so indeed, $$\boxed{\csc x+\cot x=\frac1{\csc x-\cot x}}$$

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Show backward: $$\frac{1}{\frac{1}{\sin x}-\frac{1}{\tan x}}=\frac{\sin x}{1-\cos x}=\frac{\sin x(1+\cos x)}{\sin^2 x}=\frac{1}{\sin x}+\frac{1}{\tan x}$$

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