Given an equation with trigonometric roots $\tan a$ and $\cot a$, determine $\sin a + \cos a$. Question: Given that $\tan a$ and $\cot a$ are two real roots of the equation $x^2 + k^2 - kx - 3 = 0$, and $3\pi < a < \frac{7\pi}{2}$, find the value of $\sin a + \cos a$.
My solution can be found below in the answers section.
 A: My solution:
First realize that the range for $a$ can be simplified to $\pi < a < \frac{3\pi}{2}$.
Since $\tan a$ and $\cot a$ have a product of $1$, using Vieta's formula, $k^2 - 3 = 1 \Longrightarrow k = \pm 2$.
So, either $x^2 + 2x + 1 = 0$ or $x^2 - 2x + 1 = 0$.
Hence, $x = \pm 1$.
So, using the fact that $\tan a$ and $\cot a$ have a product of $1$, either $\tan a = \cot a = 1$ or $\tan a = \cot a = -1$.
If $\tan a = \cot a = 1$ $\Longrightarrow \sin a = \cos a.$ This would make $a = \frac{\pi}{4}$. However, this does not fit within the range for $a$.
So, $\tan a = \cot a = -1 \Longrightarrow \sin a = -\cos a.$ This would make $a = \frac{5\pi}{4}$. This fits within the range for $a$.
However, we must read the problem carefully as it asks for the value of $\sin a + \cos a$ and not just the value of $a$ itself.
So, the answer is $-\frac{\sqrt{2}}{2} + -\frac{\sqrt{2}}{2} = \boxed{-\sqrt{2}}$
A: First of all, in the given range $\sin a,\cos a<0$
$\implies y=\sin a+\cos a<0$
Now $k=\tan a+\cot a=\dfrac1{\sin a\cos a}>0$
Again $k^2-3=\tan a\cot a=1\implies k=?$ as $k>0$
Now $$y^2=1+\dfrac2k$$
Finally use the fact $y<0$
