Understanding the Arctan Identity: Solving for Inverse Trigonometric Functions

[ali PMPAINT]

Homework Statement

arctan(1)+arctan(2)+arctan(3)=Pi

Relevant Equations

I think tan(a)+tab(b)+tan(c)=tan(a)*tan(b)*tan(c)

So, I saw the answer but I couldn't understand it. But I think it can be solved by tan(a)+tab(b)+tan(c)=tan(a)*tan(b)*tan(c) (where a+b+c=Pi) , but I don't know how to transfer it to its inverse.

The answer:

 

[BvU]

Hello ali, $\qquad$ $\qquad$ !

I think it can be solved by $\tan a +\tan b + \tan c =\tan a * \tan b * \tan c$ (where $a+b+c=\pi$)

Thanks for that equation -- I didn't know about it.
But: wouldn't it be easier to use some other equations, e.g. the expressions for $\ \tan (a\pm b)$ ?

 

[ali PMPAINT]

Hello ali, $\qquad$ $\qquad$ !

Thanks
But: wouldn't it be easier to use some other equations, e.g. the expressions for $\ \tan (a\pm b)$ ?
[/QUOTE]
I think it would, but I don't know how to use these expressions for arctan.
And if you understood the solution book mentioned, could you explain it to me?
and how did you use "Pi" symbol?

 

[ali PMPAINT]

But: wouldn't it be easier to use some other equations, e.g. the expressions for $\ \tan (a\pm b)$ ?
[/QUOTE]
Akshely(Sorry for my English, I don't know how to spell it correctly), yes! Thanks for your advice. for anyone interested for the proof:

 

[BvU]

how did you use "Pi" symbol?

I type ## \pi ## and get $\pi$

How to typeset equations is described in this tutorial (from point 7 in the guidelines)

solution book mentioned

On the lower right you see a triangle with a top left angle a that has tangent 1.

On top of that angle is a triangle with a lower left angle b that has tangent 2.
And c has tangent 3.

spell it correctly

that would be: actually

 

[ali PMPAINT]

I type ## \pi ## and get $\pi$
On the lower right you see a triangle with a top left angle a that has tangent 1.
View attachment 242925
On top of that angle is a triangle with a lower left angle b that has tangent 2.
And c has tangent 3.

that would be: actually

Oh! Now I got it! Thank you very much!

 

[ehild]

From the right triangles ABC, CBD, CDE, and AFE, the length of AE is the same as the length of AC + length of CE, so the line ACE is a straight line, the angle C is 180°.