showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$

I’m having problem with showing that:
$$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$

I would need some help in the right direction

Solutions Collecting From Web of "showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$"

$(3+2i)^2=5+12i$, so $\arctan\frac23+\arctan\frac23=\arctan\frac{12}{5}$.

Alternatively, study this diagram: diagram goes here

From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),

$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1\end{cases} $$

$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{ if }x^2<1$$

Here $\displaystyle x=\frac23$

There is a nice geometrical way of showing that. Your statement can be shown with the bisector theorem (… Check the picture


Sorry but I have quickly drawn it. Now your problem is equivalent to say that $\arctan(\frac{2}{3}) = \alpha$ since $\beta = \arctan(\frac{12}{5})$. For the bisector theorem it is true that
\frac{BC}{CD} = \frac{5}{13}
now we know that $BC+CD=12$ and solving the two equations we find that $BC=10/3$ that is exactly equivalent saying
\arctan(\alpha) = \frac{2}{3}
that is exactly what we wanted to demonstrate.

This was fun 😉

The basic concept of inverse trigonometry is as follows –
Arctan is same as tan^-1 with in the specified values.

tan^-1 x + tan ^-1 y = tan ^ -1 (x+ y/1−xy) if xy<1
π+ tan ^ -1 (x+y/1−xy) if xy>1 and x>0 and y>0
-π + tan ^ -1 (x + y/1-xy) if xy>1 and x<0 and y

This concept will solve your proof above by instantly putting the given value of x.