Amazing Trigonometry_1

For $\triangle ABC$ , $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ , $\implies \cot A\cot B + \cot B \cot C + \cot C \cot A = 1$ . Now we have:

$\begin{aligned} \cot A + \cot B + \cot C & = \sqrt 3 & \small \color{#3D99F6} \text{Squaring both sides} \\ \cot^2 A + \cot^2 B + \cot^2 C + 2({\color{#3D99F6}\cot A\cot B + \cot B \cot C + \cot C \cot A}) & = 3 \\ \cot^2 A + \cot^2 B + \cot^2 C + 2({\color{#3D99F6}1}) & = 3 \\ \cot^2 A + \cot^2 B + \cot^2 C & = \color{#3D99F6} 1 \\ \cot^2 A + \cot^2 B + \cot^2 C & = \color{#3D99F6} \cot A\cot B + \cot B \cot C + \cot C \cot A \end{aligned}$

The last equation above is satisfied when $A=B=C$ , hence $\triangle ABC$ is an equilateral triangle .

Let's use $\cot { A }$ = $\sqrt{3}/3$ , it appears to be fitting. The previous is also $1/\sqrt{3}$ rationalized, so we draw a triangle and see that the $1$ must be the adjacent side of the angle, and $\sqrt{3}$ the opposite side of the angle $cot A$ . By the Pythagorean theorem, the hypotenuse is 2 and with this we know that the angle $A$ is $60 degrees$ OR $\pi/3$ .

So, if a triangle has the angles $A,B,C$ each of equal length, $\pi/3$ in this case, then it is equilateral

Given $\cot { A } +\cot { B } +\cot { C }$ = $\sqrt { 3}$

OkaY now since the A,B,C are the angles of a triangle so the sum of angle of any triangle is equal to 180°

$\tan {(A+B+C)} = 0$

Therefore ,

$\tan{A} + \tan{B} + \tan{C} = \tan {A} \times \tan{B} \times \tan{C}$

We came to this result by using the formula $\tan{(A+B+C)} =\frac { \tan{A} + \tan{B} + \tan{C} - \tan {A} \times \tan{B} \times \tan{C} }{1- (\tan{A} \times \tan{B} \times \tan{C)} }$

Now , Dividing by $\tan{A} \times \tan{B} \times \tan{C}$ Both sides of equation $\tan{A} + \tan{B} + \tan{C} = \tan {A} \times \tan{B} \times \tan{C}$ We get $(\cot{A} \times \cot{B} )+( \cot {B} \times \cot{C}) + (\cot{C} \times \cot{A}) =1$

Squaring both sides of the equation

$\cot { A } +\cot { B } +\cot { C }$ = $\sqrt { 3}$ We get $\cot^2{A} + \cot^2{B} + \cot^2{C} + 2 \times ((\cot{A} \times \cot{B} )+( \cot {B} \times \cot{C}) + (\cot{C} \times \cot{A})) = 3$

Therefore , $\cot^2{A} + \cot^2{B} + \cot^2{C} =1$

Now using the inequality , $(x^2 + y^2 + z^2) \geq (xy + yz + zx)$

Now multiplying both sides by two and adding $x^2 + y^2 + z^2$

We get

$3(x^2 + y^2 + z^2) \geq (x+y+z)^2$

Substitute $x=\cot{A}$

, $y =\cot{B}$

and $z= \cot{C}$

We get $3(\cot^2{A} + \cot^2{B}+ \cot^2{C}) \geq (\cot{A}+ \cot{B} +\cot{C} )^2$

As we substitute the values of $(\cot^2{A} + \cot^2{B}+ \cot^2{C})$

and $\cot{A}+ \cot{B} +\cot{C}$

We get to know that the equality holds ,

This implies

$\cot{A} = \cot{B} = \cot{C}$

Therefore we get ,

$A= B = C = \frac {π}{3}$

So the answer is equilateral triangle.

P.S : Please share your ideas about representing the answer.

Bonus : Prove the inequality that I have used.