Using only the basic identities ($\sin^2{A}+\cos^2{A}=1$, $1+\cot^2 A=\csc^2{A}$ and $1+\tan^2 A=\sec^2{A}$) show that:

$$ \frac{1}{\csc{A}-\cot{A}}-\frac{1}{\sin A}=\frac{1}{\sin A}-\frac{1}{\csc A+\cot A} $$

I've tried a few different ways but I can't seem to show it adequately. This is the only way I can think of doing it using what I've done so far (though I suspect there is something very straightforward I've overlooked), if $x=\sin A $ and $y=\cos A$:

$$ \frac{1}{\frac{1}{x}-\frac{y}{x}}-\frac{1}{x}=\frac{x^2-1+y}{x(1-y)}=\frac{y-y^2}{x(1-y)}=\frac{y(1-y)}{x(1-y)}$$ Now cancelling the $(1-y)$ gives $\cot{A}$ so they only way I can think of showing the identity is by writing it like: $$\frac{y(1-y)(1+y)}{x(1-y)(1+y)}=\frac{y(1+y)}{x(1+y)}=\frac{y+y^2}{x(1+y)}=\frac{(y+1)-x^2}{x(y+1)}=\frac{1}{x}-\frac{x}{y+1}=\frac{1}{x}-\frac{1}{\frac{1}{x}+\frac{y}{x}} $$ which is the required result but I'm not sure whether it's strictly valid given the fact that I had to work backwards from the right hand identity to spot that if I multiplied the numerator and denominator by $(y+1)$ it would work. Plus it looks really quite a simple problem so I'm sure I'm missing something, so my question: Is my method valid and even if it is are they any easier way to show this?

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    $\begingroup$ I think you meant $\;y=\color{red}{\cos A}\;$ ... $\endgroup$ – DonAntonio Apr 5 '14 at 14:11
  • $\begingroup$ indeed, it seems that no matter how many times I reread it there's always a damn typo. $\endgroup$ – Jay Apr 5 '14 at 14:13

As $\displaystyle\csc^2A-\cot^2A=1,(\csc A+\cot A)(\csc A-\cot A)=1$

$\displaystyle\implies\frac1{\csc A-\cot A}=\csc A+\cot A$ and $\displaystyle\frac1{\csc A+\cot A}=\csc A-\cot A$

$\displaystyle\implies\frac1{\csc A-\cot A}+\frac1{\csc A+\cot A}=\csc A+\cot A+(\csc A-\cot A)=\frac2{\sin A}$

$\displaystyle\implies\frac1{\csc A-\cot A}+\frac1{\csc A+\cot A}=\frac1{\sin A}+\frac1{\sin A}$

Now rearrange the terms


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