# Can a denominator of fraction be multiplied by -1 without affecting the numerator ? and if so why?

I have been presented with a solution for solving trigonometric identities. However I would like to see further proof that one of the lines of work are valid. \begin{align*} \frac{2\sin x\cos x}{1 + (\cos x)^2 - (\sin x)^2} & = \frac{2\sin x\cos x}{1 - (\sin x)^2 + (\cos x)^2}\\ & = \frac{2\sin x\cos x}{2(\cos x)^2}\\ & = \frac{\sin x}{\cos x}\\ & = \tan x \end{align*}

As shown the denominator is multiplied by $-1$, however the numerator has not. Is this valid? And if so, why?

Any help is greatly appreciated.

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jan 30 '16 at 23:54
• Will do, Cheers ! – Flewitt Connor Jan 30 '16 at 23:55
• Great display name, RIP Futurama. – Matt Samuel Jan 31 '16 at 0:29

## 1 Answer

There is no multiplication by $-1$, only a reordering of the terms: $$1+(\cos x)^2-(\sin x)^2 = 1-(\sin x)^2+(\cos x)^2.$$

• Thanks for clearing that up, what a silly mistake ! – Flewitt Connor Jan 30 '16 at 23:50
• I've done way worse (and will most likely still do). :) – Clement C. Jan 30 '16 at 23:51