tan(A/2) + cot(A/2) = 2cosecA

Hello

Could someone please help me start with proving the following identity:

$\displaystyle \tan \dfrac{A}{2} + \cot \dfrac{A}{2} \equiv 2\csc A$

I'm trying to show that the L.H.S = the R.H.S. I've tried finding the half angle formula for $\displaystyle \tan \dfrac{A}{2}$, but it looks like a quadratic. I have also tried writing the L.H.S as a single term. I'm totally stuck. A hint would be greatly appreciated.

Thank you

Re: tan(A/2) + cot9A/2) = 2cosecA

Quote:

Originally Posted by

**Furyan** Hello

Could someone please help me start with proving the following identity:

$\displaystyle \tan \dfrac{A}{2} + \cot \dfrac{A}{2} \equiv 2\csc A$

To save notation let $\displaystyle u=\frac{A}{2}$.

$\displaystyle \frac{\sin(u)}{\cos(u)}+\frac{\cos(u)}{\sin(u)}= \frac{1}{\sin(u)\cos(u)}=\frac{2}{\sin(2u)}$

Re: tan(A/2) + cot9A/2) = 2cosecA

Hello Plato,

Thank you so much. I clearly wasn't trying hard enough to write the L.H.S as a single term. I totally missed that.

Thank you for all your help and patience, it's very much appreciated.

Re: tan(A/2) + cot9A/2) = 2cosecA

...and I actually understand the use of the double angle formula to equate the following.

$\displaystyle \frac{1}{\sin(u)\cos(u)}=\frac{2}{\sin(2u)}$

Edit: Understand might be an over statement, but I get it.