Proving an identity that's proving to be complex

**struck** · July 21st 2009, 01:16 PM

Prove the identity cot A/2 - tan A/2 = 2cot A

I have written it as:

$\frac{cos \frac{A}{2}}{sin \frac{A}{2}} - \frac { sin \frac {A}{2}}{cos \frac{A}{2}}$

Taking LCM:

$\frac {cos^2 \frac{A}{2} - sin^2 \frac{A}{2}}{sin \frac{A}{2} cos \frac{A}{2}}$

Now I don't know how to solve it any further?

Is there any easier method of proving this identity ?

**e^(i*pi)** · July 21st 2009, 01:30 PM

Originally Posted by struck

Prove the identity cot A/2 - tan A/2 = 2cot A

I have written it as:

$\frac{cos \frac{A}{2}}{sin \frac{A}{2}} - \frac { sin \frac {A}{2}}{cos \frac{A}{2}}$

Taking LCM:

$\frac {cos^2 \frac{A}{2} - sin^2 \frac{A}{2}}{sin \frac{A}{2} cos \frac{A}{2}}$

Now I don't know how to solve it any further?

Is there any easier method of proving this identity ?

I'd do it your way too and you were only two steps away from the end. Carrying on:

For the numerator recall that $cos(2u) = cos^2(u) - sin^2(u)$

For the denominator recall that $sin(2u) = 2sin(u)cos(u) \rightarrow cos(u)sin(u) = \frac{1}{2}sin(2u)$

This will reduce your equation to $\frac{cos(2u)}{\frac{1}{2}sin(2u)}$ which I'm sure you can simplify to get the answer

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