1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.