Trigonometric Proof help (1 Viewer)

gwilymprice · Mar 28, 2014

So the problem is (tan^2x - tanx)/(tan^2x + tanx) = tan2x - sec2x

I am completely stuck on it and have put it into trig solvers and they give me huge proofs, but it is only worth one mark so it heaps too complicated for the marks allocated.
Can anyone enlighten me with the way of doing it?
Thanks in advice bored of studies

rumbleroar · Mar 28, 2014

What have you tried so far?

Ikki · Mar 28, 2014

Easiest way i can think of is:

$RHS= \\ \\ \frac{2tanx}{1-tan^2x}-\frac{1}{cos^2x-sin^2x} \\ \\ $Then for the second fraction multiply top and bottom by sec^2x.$ \\ =\frac{2tanx}{1-tan^2x}-\frac{sec^2x}{1-tan^2x} \\ \\ $Common denominator$ \\ =-\frac{tan^2x-2tanx+1}{(1-tanx)(1+tanx)} \\ \\ $Now factorise top and cancel with bottom.$ \\ \therefore \frac{tanx-1}{tanx+1}\\ \\=LHS$

gwilymprice · Mar 28, 2014

Okay that is 1000 times easier than what I was doing! I always fall into the trap of doing LHS first everytime!
Thank you so much

Trigonometric Proof help (1 Viewer)

gwilymprice

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rumbleroar

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Ikki

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gwilymprice

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