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The expression is $\dfrac 1 {(1-\csc x) (1+\csc x)}$. The identities I'm working with are:

$$\sin^2 x + \cos^2 x = 1 \\ \frac {\sin x} {\cos x} = \tan x \\ \sec^2 x = 1+\tan^2x \\ \csc^2 x = 1+\cot^2x .$$

I've tried pretty much everything I can to simplify it, but to no avail. This is not homework, simply a question that cropped up while revising. Thank you!

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  • $\begingroup$ Is (1-cosecx)(1+cosecx) both in the denominator? $\endgroup$ – Ralff Nov 6 '16 at 19:14
  • $\begingroup$ Yes, they're both in the denominator $\endgroup$ – The White Wolf Nov 6 '16 at 19:15
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One way could be $$\frac { 1 }{ \left( 1-\csc { x } \right) \left( 1+\csc { x } \right) } =\frac { 1 }{ 1-\csc ^{ 2 }{ x } } =\frac { 1 }{ 1-\frac { 1 }{ \sin ^{ 2 }{ x } } } =\frac { \sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } -1 } =-\frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } =-\tan ^{ 2 }{ x } $$

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  • $\begingroup$ Thanks, perfect. I overlooked multiplying both the numerator and denominator by sin^2x $\endgroup$ – The White Wolf Nov 6 '16 at 19:27

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