Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$

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 $x \in (45^{\circ},90^{\circ})$

$\sin \alpha + \cos \alpha$ 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.