$\arctan(2)+\arctan(3)+\arctan(4)=?$

Question

This question came in the Agricultural universities' cluster exam 20-21

Q) If $\arctan(2)+\arctan(3)+\arctan(4)=\theta$ then $\tan\theta=?$

(a) 9

(b) 7/2

(c) 3/5

(d) 4/5

Third-party question bank's attempt:

$$\theta=\arctan(2)+\arctan(3)+\arctan(4)=\arctan\frac{2+3+4-2\cdot3\cdot4}{1-2\cdot3-3\cdot4-4\cdot2}$$

$$\theta=\arctan\frac{3}{5}$$

$$\tan\theta=\frac{3}{5}$$

So, (c).

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The formula that the question bank used is

$$\arctan(x)+\arctan(y)+\arctan(z)=\arctan\frac{x+y+z-xyz}{1-xy-yz-xz}$$

$$\text{where $xy+yz+zx\leq 1$}$$

However, $xy+yz+zx$ in the case of the above question is

$$xy+yz+zx=2\cdot3+3\cdot4+4\cdot2=26\nleq 1$$

So, isn't the third-party question bank's attempt wrong?

Answer 1

Alternative Solution:

Let $\alpha = \arctan2$, $\beta=\arctan3$, $\gamma=\arctan4$

then $\tan\alpha =2$,$\tan\beta =3$, $\tan\gamma =4$, where $\theta=\alpha+\beta+\gamma$

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha \times \tan\beta}=\frac{5}{-5}=-1$

$\tan\theta = \tan[(\alpha+\beta)+\gamma]=\frac{4+(-1)}{1-4\times(-1)}=\frac{3}{5}$

Answer 2

For every $a, b\in \mathbb{R}$ with $ab\ne 1$ we have $$ \tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a))+\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}=\frac{a+b}{1-ab} $$

so $$\tag{1} \arctan(a)+\arctan(b)=\arctan\left(\frac{a+b}{1-ab}\right) \quad \forall a, b \in \mathbb{R}, \quad ab\ne 1. $$ Since $\displaystyle 2\cdot 3, \frac{2+3}{1-2\cdot3} \ne 1$, we have

\begin{eqnarray} \arctan(2)+\arctan(3)+\arctan(4) &=& \arctan\left(\frac{2+3}{1-2\cdot3}\right)+\arctan(4)\cr &=&\arctan(-1)+\arctan(4)\cr &=&\arctan\left(\frac{-1+4}{1-(-1\cdot4)}\right)\cr &=&\arctan\left(\frac35\right) \end{eqnarray}

Answer 3

arctan is a little unpredictable; I suggest taking $A, B,C $ which are going to be the three arctans

Then name $ \tan A = x, \tan B = y, \tan C = z.$ As others have said, $$ \tan (A+B) = \frac{x+y}{1-xy} $$ and then $$ \tan (A+B+C) = \frac{x+y+z - xyz}{1 - xy - yz - zx} $$ where the three Vieta combinations of three roots of some cubic are visible. With $x=2, y=3, z=4$ this becomes $$ \tan (A+B+C) = \frac{2+3+4 - 2 \cdot 3 \cdot 4}{1 - 2 \cdot 3 - 3 \cdot 4 - 4 \cdot 2} = \frac{9-24}{1-26} = \frac{-15}{-25} = \frac{3}{5} $$ Next, in what quadrant is the angle $A+B+C?$ In degrees, we know $\tan 60^\circ = \sqrt 3$ and this is smaller than $2.$ Thus the three arctans are above $60^\circ$ but below $90^\circ.$ Tus their sum is bigger than $180^\circ$ and below $270^\circ,$ therefore in the third quadrant. Thus $A+B+C = 180^\circ + \arctan \frac{3}{5} $ in degrees or radians (real numbers) $$A+B+C = \pi + \arctan \frac{3}{5}.$$

$$\arctan 2 + \arctan 3 + \arctan 4 = \pi + \arctan \frac{3}{5}.$$ Again, note that $ \pi + \arctan \frac{3}{5} $ is not arctan of anything.