By multiplying by \(\cot \alpha \) on top and bottom of the fraction, we have \begin {align*} f_{\alpha }(x) & = \arctan \left (\frac {x + \cot \alpha }{1 - x \cot \alpha }\right ) \\ & = \arctan \left (\frac {x + \tan \left (\frac {\pi }{2} - \alpha \right )}{1 - x \tan \left (\frac {\pi }{2} - \alpha \right )}\right ) \\ & = \arctan \tan \left (\arctan x + \frac {\pi }{2} - \alpha \right ). \end {align*}
Since \(\arctan x \in \left (-\frac {\pi }{2}, \alpha \right ) \cup \left (\alpha , \frac {\pi }{2}\right )\), we have \[ \arctan x + \frac {\pi }{2} - \alpha \in \left (-\alpha , \frac {\pi }{2}\right ) \cup \left (\frac {\pi }{2}, \pi - \alpha \right ). \]
Hence, we can simplify this to \begin {align*} f_{\alpha }(x) & = \arctan \tan \left (\arctan x + \frac {\pi }{2} - \alpha \right ) \\ & = \begin {cases} \arctan x + \frac {\pi }{2} - \alpha , & x < \tan \alpha , \\ \arctan x - \frac {\pi }{2} - \alpha , & x > \tan \alpha . \end {cases} \end {align*}
Hence, by differentiating with respect to \(x\), the constants differentiate to \(0\), and hence \begin {align*} f'_{\alpha }(x) & = \DiffOp {x} \arctan x \\ & = \frac {1}{1 + x^2}, \end {align*}
as desired.
The graph consists of \(2\) branches of \(\arctan \), as the simplified expressions suggests. We have the following limiting behaviours of \(f_{\alpha }\): \begin {align*} \lim _{x \to -\infty } f_{\alpha } (x) & = \lim _{x \to -\infty } \arctan x + \frac {\pi }{2} - \alpha = -\alpha , \\ \lim _{x \to \tan \alpha ^{-}} f_{\alpha } (x) & = \frac {\pi }{2}, \\ \lim _{x \to \tan \alpha ^{+}} f_{\alpha } (x) & = -\frac {\pi }{2}, \\ \lim _{x \to \infty } f_{\alpha } (x) & = \lim _{x \to \infty } \arctan x - \frac {\pi }{2} - \alpha = -\alpha , \end {align*}
which shows that \(f_{\alpha }\) has a horizontal asymptote with equation \(y = -\alpha \).
For the intersection with the \(y\)-axis, \[ f_{\alpha } (0) = \arctan 0 + \frac {\pi }{2} - \alpha = \frac {\pi }{2} - \alpha , \] and for the intersection with the \(x\)-axis, \[ f_{\alpha } (x) = 0 \iff x \tan \alpha + 1 = 0\iff x = - \cot \alpha . \]
The graph looks as follows.
The domain of this new graph is \(x \in \RR \setminus \{\tan \alpha , \tan \beta \}\). By considering the functions in the different corresponding ranges, we have \[ f_{\alpha } (x) - f_{\beta }(x) = \begin {cases} \left (\arctan (x) + \frac {\pi }{2} - \alpha \right ) - \left (\arctan (x) + \frac {\pi }{2} - \beta \right ) = \beta - \alpha , & x < \tan \alpha , \\ \left (\arctan (x) - \frac {\pi }{2} - \alpha \right ) - \left (\arctan (x) + \frac {\pi }{2} - \beta \right ) = \beta - \alpha - \pi , & \tan \alpha < x < \tan \beta , \\ \left (\arctan (x) - \frac {\pi }{2} - \alpha \right ) - \left (\arctan (x) - \frac {\pi }{2} - \beta \right ) = \beta - \alpha , & \tan \beta < x. \end {cases} \]
Hence, the graph looks as follows.
By differentiation, we have \begin {align*} g'(x) & = \frac {1}{1 - \sin ^2 x} \cos x - \frac {1}{\sqrt {1 + \tan ^2 x}} \sec ^2 x \\ & = \frac {\cos x}{\cos ^2 x} - \frac {\sec ^2 x}{\abs *{\sec x}} \\ & = \sec x - \abs *{\sec x} \\ & = \begin {cases} \sec x - \sec x = 0, & 0 \leq x < \frac {1}{2}\pi \text { or } \frac {3}{2} \pi < x \leq 2\pi , \\ \sec x - (- \sec x) = 2 \sec x, & \frac {1}{2}\pi < x < \frac {3}{2}\pi , \end {cases} \end {align*}
since \(\sec x\) takes the same sign as \(\cos x\), which is negative when \(\frac {1}{2}\pi < x < \frac {3}{2}\pi \), and positive when \(0 \leq x < \frac {1}{2}\pi \) or \(\frac {3}{2} \pi < x \leq 2\pi \) within the range.
For \(\frac {1}{2} \pi < x < \frac {3}{2}\pi \), we must have \[ g(x) = \ln \abs *{\tan x + \sec x} + C = \ln \left (- \tan x - \sec x\right ) + C, \] and by verifying \[ g(\pi ) = \artanh (0) - \arsinh (0) = 0, \] we can see \(C = 0\).
Hence, for \(0 \leq x < \frac {1}{2}\pi \) and \(\frac {3}{2} \pi < x \leq 2\pi \) respectively, \(g(x)\) is constant, and notice that \[ g(0) = g(2\pi ) = 0, \] and hence \[ g(x) = \begin {cases} \ln \left (- \tan x - \sec x\right ), & \frac {1}{2} \pi < x < \frac {3}{2}\pi , \\ 0, & 0 \leq x < \frac {1}{2}\pi \text { or } \frac {3}{2} \pi \leq 2\pi . \end {cases} \]
