Triple Tangents

Given that $(\tan A + \tan B)^2 = \tan^2 A + 2\tan A \tan B + \tan^2 B = 19 \ \ ...(1)$ and $\implies \tan A + \tan B = \sqrt{19}$ . Also

$\begin{aligned} \tan^3 A + \tan^3 B & = 47 \\ (\tan A + \tan B)(\tan^2 A - \tan A \tan B + \tan^2 B) & = 47 \\ \implies \tan^2 A - \tan A \tan B + \tan^2 B & = \frac {47}{\sqrt{19}} & ...(2) \end{aligned}$

Then $(1)-(2): \ \ 3 \tan A \tan B = 19-\dfrac {47}{\sqrt{19}} \implies \tan A \tan B = \dfrac {19\sqrt{19}-47}{3\sqrt{19}}$ . And we have:

$\begin{aligned} \tan C & = \tan (\pi - A - B) = - \tan (A+B) = - \frac {\tan A + \tan B}{1-\tan A\tan B} = \frac {\sqrt{19}}{\frac {19\sqrt{19}-47}{3\sqrt{19}}-1} \approx 2.506333648 \\ \implies \angle C & \approx 1.191161652 \text{ rad} \end{aligned}$

Therefore $\cot \dfrac C2 + \tan C - \csc C + 0.071 \approx \boxed{2.976}$ .

Let $\Omega = \tan C$ .

We have, $(\tan A + \tan B)^ {3} = \tan^ {3} A + \tan^ {3} B + 3 \tan A \cdot \tan B(\tan A + \tan B$ ) $=$ $19^ {\frac {3}{2}}$ .

Substituting values and solving for $(\tan A \cdot \tan B)$ gives us $\tan A \cdot \tan B = 2.739$ . Also, we know that $A + B+ C = \pi$ . So, by the triple tangent identity (see below) , $\tan A+ \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$ . Therefore,

$\Omega + \sqrt {19} = 2.739 \times \Omega$ $\Rightarrow$ $\Omega = \tan C = \boxed {2.506}$ . Now, using the half angle formulas (see below), $\cot (\frac {C}{2}) = \sqrt {\frac {1 + \cos C}{1 - \cos C}} = \frac {1 + \cos C}{\sin C} = \csc C + \cot C$ .

So we can conclude that, $\cot (\frac {C}{2}) - \csc C + \tan C + 0.071 = \cot C + \tan C + 0.071 = \frac {1}{2.506} + 2.506 + 0.071 = \boxed {2.976}$

References :

Triple tangent identity

Half angle formulas