# Thread: Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. angle

1. ## Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. angle

Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using double angle

I do not know where to begin with this one. Any help would be appreciated.

2. Originally Posted by smsmith
I do not know where to begin with this one.
Maybe... use the double-angle identities...?

Eliz.

3. First of all it has to be $\left[ {\tan (x)} \right]^2 = \tan ^2 (x).$
$\begin{array}{rcl} \frac{{1 - \cos (2x)}}{{1 + \cos (2x)}} & = & \frac{{1 - \left[ {1 - 2\sin ^2 (x)} \right]}}{{1 + \left[ {2\cos ^2 (x) - 1} \right]}} \\ & = & \frac{{2\sin ^2 (x)}}{{2\cos ^2 (x)}} \\ \end{array}$

4. ## Re: Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. an

Hello, smsmith!

Verify: $\,\tan^2x \:= \:\frac{1\,-\,\cos2x}{1\,+\,\cos2x}$ using double-angle identities.

You're expected to know these identities:

. . $\sin^2x \:=\:\frac{1\,-\,\cos2x}{2}\;\;\;\;\;\cos^2x \:=\:\frac{1\,+\,\cos2x}{2}$

We have: $\:\tan^2x \:=\:\L\left(\frac{\sin x}{\cos x}\right)^2\:=\:\frac{\sin^2x}{\cos^2x} \:=\:\frac{\frac{1\,-\,\cos2x}{2}}{\frac{1\,+\,\cos2x}{2}} \:=\:\frac{1\,-\,\cos2x}{1\,+\,\cos2x}$

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