Quick question,

I'm proving an identity, and I want to know if -tanx is the same as 1-tanx?

I feel as though they are, someone please clarify!

Actually, while I'm at it, could you see if I've proven this identity correctly?

\(prove\frac{\cos ^2x-\sin ^2x}{\cos ^2x+\sin \left(x\right)\cos \left(x\right)}=1-\tan \left(x\right)\)

Guest May 26, 2017

#1**+1 **

Hi

\(\frac{cos^2(x)-sin^2(x)}{cos^2(x)+sin(x)cos(x)}=1-tan(x)\\~\\ LHS=\frac{cos^2(x)-sin^2(x)}{cos^2(x)+sin(x)cos(x)}\\ LHS=\frac{(cos(x)-sin(x))(cos(x)+sin(x))}{cos(x)(cos(x)+sin(x))}\\ LHS=\frac{cos(x)-sin(x)}{cos(x)}\\ LHS=1-tan(x)\\ LHS=RHS\qquad \qquad QED\)

Here are the 2 graphs that you talke about. You can see they are not the same.

Melody
May 26, 2017