Trigonometric exercise asked in the Pretty mathematics by Kunny Shogun

Exercise:

Let $A,B$ and$C$ be positive acute angles such that $\tan A=2\,\,,\,\,\tan B=4\,\,\,\And \,\,\tan C=13$

Determine the value of $A+B+C$

Solution: we know that $\tan \left( x+y \right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$


So $\tan \left( A+B+C \right)=\tan \left( A+\left( B+C \right) \right)=\frac{\tan \left( A \right)+\tan \left( B+C \right)}{1-\tan A\tan \left( B+C \right)}$

$\Rightarrow \tan \left( A+B+C \right)=\frac{\tan A+\frac{\tan B+\tan C}{1-\tan B\tan C}}{1-\tan A\frac{\tan B+\tan C}{1-\tan B\tan C}}=\frac{\frac{\tan A\left( 1-\tan B\tan C \right)+\tan B+\tan C}{1-\tan B\tan C}}{\frac{1-\tan B\tan C-\tan A\left( \tan B+\tan C \right)}{1-\tan B\tan C}}$

$\Leftrightarrow \tan \left( A+B+C \right)=\frac{\tan A-\tan A\tan B\tan C+\tan B+\tan C}{1-\tan B\tan C-\tan A\tan B-\tan A\tan C}$

$\Leftrightarrow \tan \left( A+B+C \right)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\left( \tan A\tan B+\tan A\tan C+\tan B\tan C \right)}$

$\Leftrightarrow \tan \left( A+B+C \right)=\frac{19-104}{1-\left( 8+26+52 \right)}=\frac{-85}{1-86}=\frac{-85}{-85}=1$

So $A+B+C=\arctan \left( 1 \right)=\frac{\pi }{4}\,\,or\,\,\frac{5\pi }{4}$

But $A=\arctan \left( 2 \right)>\frac{\pi }{4}$ , $B=\arctan \left( 4 \right)>\frac{\pi }{4}\,\,\And \,\,C=\arctan \left( 13 \right)>\frac{\pi }{4}$

So $A+B+C>\frac{3\pi }{4}$ thus the accepted value is $A+B+C=\frac{5\pi }{4}$