(1+cosA)^2-(1-cosA)^2-sin^2=ctgA*sinA*cosA

makarov1901

1. The problem statement, all variables and given/known data
(1+cosA)^2-(1-cosA)^2-sin^2A=ctgA*sinA*cosA


2. Relevant equations



3. The attempt at a solution

I moved sin^2 to the right side, then expanded the left side and got to:
1+2cosA+cos^2A-1+2cosA-cos^2A=1
When I cancel the left side I get:
4cosA=1

I'd be very grateful if anyone help me.

vela

Looks good so far. What's the problem?

(I assume when you wrote sin^2, you meant sin² a.)

makarov1901

You assumed right.
The problem is that both parts of the identity should be equal.
As far as I know, 4cosA is not equal to 1.

vela

It's not an identity. For example, when a=pi/2, the lefthand side is -1 while the righthand side equals 0.

Perhaps there's a typo in the problem.

makarov1901

My task is to transform both parts so that they're equal. I don't know why you say it's not an identity.

vela

If it were an identity, it would hold for all values of a. It clearly doesn't; therefore, it's not an identity.