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.
 A: $\sin(\alpha)-\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}-\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(-\sin(\alpha)\cos\left(\frac{3\pi}{4}\right)-\cos(\alpha)\sin\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)$
$\sin(\alpha)+\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}+\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(\sin(\alpha)\sin\left(\frac{3\pi}{4}\right)-\cos(\alpha)\cos\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\cos\left(\alpha+\frac{3\pi}{4}\right)$
Finally,
$$\frac{\left(\sin(\alpha)-\cos(\alpha)\right)^2}{\left(\sin(\alpha)+\cos(\alpha)\right)^2}=\frac{\left(-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)\right)^2}{\left(-\sqrt{2}\cos\left(\alpha+\frac{3\pi}{4}\right)\right)^2}=\tan^2\left(\alpha+\frac{3\pi}{4}\right)$$
A: $$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\dfrac{(\sin(\dfrac{3\pi}{4}+\alpha))^2}{(\cos(\dfrac{3\pi}{4}+\alpha))^2}=\dfrac{(\sin\dfrac{3\pi}{4}\cos\alpha+\cos\dfrac{3\pi}{4}\sin\alpha)^2}{(\cos\dfrac{3\pi}{4}\cos\alpha-\sin\dfrac{3\pi}{4}\sin\alpha)^2}=\dfrac{(\dfrac{1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)^2}{(\dfrac{-1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)^2}=\dfrac{\dfrac{1}{2}\cos^2\alpha+\dfrac{1}{2}\sin^2\alpha-\sin\alpha\cos\alpha}{\dfrac{1}{2}\cos^2\alpha+\dfrac{1}{2}\sin^2\alpha+\sin\alpha\cos\alpha}=\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$$
A: $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha} = \left(\frac{\tan \alpha-1}{\tan \alpha+1}\right)^2= \left(\frac{1-\tan \alpha}{1+\tan \alpha}\right)^2$$
$$=\tan^2\left(\dfrac{\pi}{4}-\alpha\right) =\tan^2\left(\alpha-\dfrac{\pi}{4}\right)$$
Inverse tangent function satisfies two angles in range $0, 2 \pi.\;$We are allowed to add $k\cdot\pi$ to any angle for tangent function value to be same. So for $k=1,$
$$  =\tan^2\left(\alpha-\dfrac{\pi}{4}+\pi \right) =\tan^2\left(\alpha+\dfrac{3\pi}{4}\right)= RHS $$
A: As an alternative by half-angle formula $\tan \frac{\theta}2=\frac{1-\cos \theta}{\sin \theta}$ we have (note that $\cos \alpha \neq \sin \alpha$):
$$\tan\left(\dfrac{3\pi}{4}+\alpha\right)
=\frac{1-\cos \left(\frac{3\pi}{2}+2\alpha\right)}{\sin \left(\frac{3\pi}{2}+2\alpha\right) }
=\frac{1-\sin(2\alpha)}{-\cos(2\alpha)}=\frac{(\cos \alpha-\sin \alpha)^2}{-(\cos^2 \alpha-\sin^2 \alpha)}=\frac{\cos \alpha-\sin \alpha}{-(\cos \alpha+\sin \alpha)}$$
and then
$$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{(\cos \alpha-\sin \alpha)^2}{(\cos \alpha+\sin \alpha)^2}=\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$$
A: Here’s yet another way:
$$LHS=\frac{1-\cos(\frac{\pi}{2}-2\alpha)}{1+\cos(\frac{\pi}{2}-2\alpha)}$$
$$=\frac{2\sin^2(\frac{\pi}{4}-\alpha)}{2\cos^2(\frac{\pi}{4}-\alpha)}$$
$$=\tan^2(\frac{\pi}{4}-\alpha)$$
$$=\tan^2(\alpha-\frac{\pi}{4}+\pi)$$
$$=RHS$$
A: $$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{\sin^2\left(\dfrac{3\pi}{4}+\alpha\right)}{\cos^2\left(\dfrac{3\pi}{4}+\alpha\right)}=\frac{\left(\frac{-\sqrt{2}}{2}\cos\alpha+\frac{\sqrt{2}}{2}\sin\alpha\right)^2}{\left(\frac{-\sqrt{2}}{2}\cos\alpha-\frac{\sqrt{2}}{2}\sin \alpha\right)^2}=\frac{\frac{1}{2}\cos^2\alpha+\frac{1}{2}\sin^2 \alpha-\frac{1}{2}\sin{\alpha}\cos{\alpha}}{\frac{1}{2}\cos^2 \alpha+\frac{1}{2}\sin^2\alpha+\frac{1}{2}\sin{\alpha}\cos{\alpha}}=\frac{1-\sin{2\alpha}}{1+\sin{2\alpha}}$$ Use the addition formulas for trigonometric functions.
A: 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.
A: $$
\begin{aligned}
\tan^2 \left(\frac{3 \pi}{4}+\alpha\right) &=\tan ^2\left(\pi-\frac{\pi}{4}+\alpha\right) \\
&=\tan ^2\left(\frac{\pi}{4}-\alpha\right) \\
&=\left[\frac{\sin \left(\frac{\pi}{4}-\alpha\right)}{\cos \left(\frac{\pi}{4}-\alpha\right)}\right]^2 \\
&=\left(\frac{\sin \alpha-\cos \alpha}{\cos \alpha+\sin \alpha}\right)^2 \\
&=\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}
\end{aligned}
$$
