Determine value of $m$ Consider $m = \tan x + \frac{\sin x + \cos x}{\sin x - \cos x}$ . Now determine $m$ so that expression has solution .
I used Wolfram to find range of $f(x) = \tan x + \frac{\sin x + \cos x}{\sin x - \cos x}$ but it was unable to find it!
 A: $$m=\tan x+\frac{\tan x+1}{\tan x-1}\implies m\tan x-m=\tan^2x-\tan x+\tan x+1\implies$$
$$\tan^2x-m\tan x+m+1=0$$
The discriminant of this quadratic in $\;\tan x\;$ must be non-negative, so it must be
$$\Delta=m^2-4m-4\ge0\iff\;\ldots$$
and we're done since $\;\tan x\;$ is onto $\;\Bbb R\;$ .
A: $m = \frac{\sin x}{\cos x} + \frac{\sin x + \cos x}{\sin x - \cos x}$
$m = \frac{\sin x(\sin x - \cos x) +\cos x(\sin x + \cos z)}{\cos x(\sin x - \cos x)}$
$m = \frac{\sin^{2}x + \cos^{2}x}{\sin x \cos x - \cos^{2}x}$
$(\sin x \cos x - \cos^{2}x)m - 1 = 0$
$(\tan x - 1)m - (1+\tan^{2}x) = 0$
$tan^{2}x - m\tan x +m+1 = 0$
This has real solutions when the discriminant is greater than or equal to $0$:
$m^{2} - 4(m+1) \geq 0 \Leftrightarrow (m+2)^{2} \geq 0$
So it has solutions $\forall m \in \mathbb{R}$
The range is $\tan x \in \mathbb{R}$
A: You can choose an arbitrary x value, such as $\pi$, and see what the expression gives. For x=$\pi$, this gives -1, so we know that for an m value of -1, the expression will have a solution. 
