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I have tried it several times and I do not know what I am doing wrong. This is the identity:

$$\csc^2x - \csc x = \frac{\cot^2 x}{1 + \sin x}$$

Trying with RHS, I get: $$\frac{\cos^2 x }{ \sin^2 x} + \frac{\cos^2 x }{ \sin x}$$

With LHS: $$\frac{1 }{ \sin^2 x} - \frac{\sin x }{ \sin^2 x}$$

But I do not know how to continue after that. Am I missing something?

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  • $\begingroup$ Your mistake is the expression you've derived for the RHS. $\endgroup$ – Spock Jan 19 '14 at 23:46
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Try working with the left side instead. We have

$$\csc^2 x-\csc x = \csc x(\csc x - 1) = \csc x\left(\frac{1}{\sin x}-1\right).$$

Making a common denominator, we have

$$\csc^2 x-\csc x = \csc x\frac{1-\sin x}{\sin x}.$$

We want $\frac{1}{1+\sin x}$ somehow (based on the right side), so maybe multiplying and dividing by $1+\sin x$ might do the trick.. (You can do this since multiplying by $\frac{1+\sin x}{1+\sin x}$ does not change anything..)

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$$\frac{1}{sin^2x}-\frac{1}{sinx}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}-\frac{sinx}{sin^2x}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}(1-{sinx})=\frac{1}{sin^2x}(\frac{cos^2x}{1+sinx})$$
$$\frac{1}{sin^2x}(1-{sinx})(1+sinx)=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(1-{sin^2x})=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(cos^2x)=\frac{1}{sin^2x}{cos^2x}$$
And we're done.

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    $\begingroup$ I thought that you were trained not to feel, but here I can see your secret love with the $\frac{1}{\sin^2x}$ term... $\endgroup$ – chubakueno Jan 19 '14 at 23:46
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    $\begingroup$ That's true, but I'm taking care of $sinx=0$. While I had my training at Vulcan academy, my mother is from Earth so there are feelings sometimes :-). $\endgroup$ – Spock Jan 19 '14 at 23:49
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    $\begingroup$ But, mister Spock, if $\sin x=0$ all those equations are meaningless! You designed this test, don't you remember? The equation was developed in order to test the mathematician's ability to reason in the worst conditions possible, and see how he could manage the terror. But everything breaks down when you pull the $x=n\pi$ trigger, and there is no way out to escape of the Klingon. $\endgroup$ – chubakueno Jan 20 '14 at 2:34
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We have $$\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x)$$

$$\implies \csc^2x-\csc x=\frac{1-\sin x}{\sin^2x}=\frac{\cos^2x}{\sin^2x(1+\sin x)}=\cdots$$

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