Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For  the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.

One more (why not?):
$$ \cos \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ -\frac{\sqrt2}{2} ·(\sin \alpha \ + \ \cos \alpha) $$ $$ \Rightarrow \ \ \sec \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{-\sqrt2}{(\sin \alpha \ + \ \cos \alpha)} \ \ ; $$
$$ \tan^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \sec^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ - \ 1 \ \ = \ \  \frac{2}{(\sin \alpha \ + \ \cos \alpha)^2} \ - \ 1 $$
$$ = \ \  \frac{2 \ - \ \sin^2 \alpha \ - \ 2 \sin \alpha \cos \alpha \ - \ \cos^2 \alpha}{\sin^2 \alpha \ + \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha} \ \ = \ \  \frac{1 \  - \ 2 \sin \alpha \cos \alpha }{1 \ + \ 2 \sin \alpha \cos \alpha } \ \ = \ \ \frac{1 \  - \  \sin (2\alpha)  }{1 \ + \  \sin (2\alpha) } \ \ . $$
Your approach would be a "time-reversal" of this, but I don't think it's at all obvious that you would want to replace $ \ \sin^2 \alpha + \cos^2 \alpha \ $ with $ \ 2 - \sin^2 \alpha  - \cos^2 \alpha \ $ in your numerator, and then work toward a "secant-squared" expression. (eMathHelp probably has the best way to continue from where you left off.)
ADDENDUM --
We may also apply the "angle-addition" formula for tangent:
$$ \tan  \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{\tan \alpha \ + \ \tan \frac{3 \pi}{4}}{1 \ - \ \tan \alpha · \tan \frac{3 \pi}{4}} \ \ = \ \ \frac{\tan \alpha \ + \ (-1)}{1 \ - \ \tan \alpha · (-1)} $$
$$ \Rightarrow \ \ \tan^2  \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{ \tan^2 \alpha \ - \ 2 · \tan \alpha \ + \ 1  }{\tan^2 \alpha \ + \ 2 · \tan \alpha \ + \ 1 }     $$
[multiplying the numerator and denominator by $ \ \cos^2 \alpha \ \ ] $
$$ = \ \ \frac{ \sin^2 \alpha \ - \ 2   \sin \alpha \cos \alpha \ + \ \cos^2 \alpha  }{\sin^2 \alpha \ + \ 2   \sin \alpha \cos \alpha \ + \ \cos^2 \alpha  }  \ \ ,   $$
and so to the left-side expression.
