Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$
Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?
 A: By the definition of cosine as the x-coordinate of a circle, cosine is negative in the interval $(90^{\circ},180^{\circ})$
$\cos  2x =2\cos^2x-1=1-2\sin^2x$
$\cos  2x =(\sqrt{2} \cos x-1)(\sqrt{2} \cos x+1)=-(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1)$
Using the fact that $\cos 2x$ is -ve in the interval $x \in (45^{\circ},90^{\circ})$
Prove the fact that $\cos x < \frac{1}{\sqrt{2}}< \sin x$ in the interval $x \in (45^{\circ},90^{\circ})$
Proof 2:
There is 1 more way people define $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$
Let the angles be $x,90-x,90$
Use the fact that the side opposite to the greater angle is greater.
Therefore, adjacent < opposite (for $x \in (45^{\circ},90^{\circ})$)
Therefore, $\cos x < \sin x$ for $x \in (45^{\circ},90^{\circ})$
A: $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are both positive in the given domain, so their quotient is also positive, and $|x| = x$ when $x \in \mathbb R^+$. $\sin \alpha + \cos \alpha > 0$ as $\sin \alpha, \cos \alpha$ are positive in the domain. Hence we can focus our attention to just proving $\sin \alpha > \cos \alpha$.
Divide both sides by $\cos \alpha$. Since it is positive in the given interval, the inequality does not change sign.
Thus you have $\tan \alpha > 1$, which is true because $\tan 45º = 1$ and $\tan x$ is a strictly increasing function in the given range $(45º, 90º)$. This is already sufficient as a justification, but on some insight as to why this is, $\tan x = \frac{\sin x}{\cos x}$, and $\sin x$ is increasing while $\cos x$ is decreasing in the given interval, which both increase the value of the function. Anything more rigorous has to involve a geometric argument with the unit circle or calculus.
A: Apparently your second question hasn't been answered, about what to do next. Here's what to do:
$$\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\frac{\frac{1}{\cos \alpha}}{\frac{1}{\cos \alpha}}\times\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\dots$$
I hope that's helpful.
If you need any more help please don't hesitate to ask.
A: After your 3rd line, you are actually done, without realizing it; you simply went down the wrong path.
The RHS
$$= \frac{\sin(a) + \cos(a)}{\sin(a) - \cos(a)}.$$
Edit
See the comments following this answer. 
A case can be made that my analyis is flawed, since I didn't bother to prove that the denominator above is always positive in the interval $(45^\circ, 90^\circ)$.
Addendum
Responding to

You can prove what the OP wants using pure geometry. Only using the definition of cosine and sine.

Okay: Imagine that you have a unit circle (i.e. of radius $= 1$) centered at the origin, with any point $(x,y)$ that is on the unit circle representing $(\cos[a],\sin[a])$, where $a$ is the angle formed by the two line segments $\overline{(0,0),(1,0)}$ and $\overline{(0,0),(x,y)}$.
Clearly, at $a = 45^\circ$, you have that $(x,y) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),$ since in a $45^\circ - 45^\circ - 90^\circ$ triangle, the two legs are equal.
Further, you can see geometrically, that for $a$ equal to any angle in $(45^\circ,90^\circ)$, for the corresponding point on the unit circle, the corresponding $y$ coordinate will have increased from $\frac{1}{\sqrt{2}}$, and the corresponding $x$ coordinate will have decreased from $\frac{1}{\sqrt{2}}$.
Therefore, since the $y$ coordinate represents $\sin(a)$ and the $x$ coordinate represents $\cos(a)$, you have that
$\sin(a) > \frac{1}{\sqrt{2}} > \cos(a),$ for $a$ equal to any angle in $(45^\circ,90^\circ)$.

A variation on the above argument is to notice that for $a$ equal to any angle in $(45^\circ,90^\circ)$, the slope of the line segment $\overline{(0,0),(x,y)}$ is $> 1$, which (alternatively) implies that $y = \sin(a) > x = \cos(a).$
