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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.

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jan 30 '16 at 23:54
  • $\begingroup$ Will do, Cheers ! $\endgroup$ – Flewitt Connor Jan 30 '16 at 23:55
  • $\begingroup$ Great display name, RIP Futurama. $\endgroup$ – Matt Samuel Jan 31 '16 at 0:29
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There is no multiplication by $-1$, only a reordering of the terms: $$ 1+(\cos x)^2-(\sin x)^2 = 1-(\sin x)^2+(\cos x)^2. $$

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

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