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

**Furyan** · January 20th 2013, 12:02 PM

Hello

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

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

I'm trying to show that the L.H.S = the R.H.S. I've tried finding the half angle formula for $\tan \dfrac{A}{2}$ , 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

**Plato** · January 20th 2013, 12:18 PM

Originally Posted by Furyan

Hello

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

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

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

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

**Furyan** · January 20th 2013, 12:28 PM

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.

**Furyan** · January 20th 2013, 12:45 PM

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

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

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

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

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