Be careful to the learnt formulas...

Assume $\theta=tan^{-1}1+tan^{-1}2+tan^{-1}3$

Taking $tan$ both sides,

$tan\theta=\dfrac{1+2+3-(1×2×3)}{1-(1×2+2×3+3×1)}=0$

So we can say that $\theta=n\pi$

Now we will use estimations to find out the integer n.

We know

• $tan^{-1}1=\frac{\pi}{4},$
• $\frac{\pi}{2}>tan^{-1}2>\frac{\pi}{4}$
• $\frac{\pi}{2}>tan^{-1}3>\frac{\pi}{4}$

Now we can estimate $\theta$ as $\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{5\pi}{4}>\theta>\frac{\pi}{4}+\frac{\pi}{4}+\frac{\pi}{4}=\frac{3\pi}{4}$

So possible value of n can be $1$ only, proving $\theta=\pi$

See the image below; The formula is IMPORTANT:

See the Three Square Problem .

$\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \frac{3 \pi}{2} - (\alpha + \beta + \gamma) = \boxed{\pi}$

(Angles $\alpha, \beta, \gamma$ are defined in the video.)

tan^-1(1) + tan^-1(2) + tan^-1(3) = pi/4 + tan^-1(2) + tan^-1(3) \Rightarrow pi/4 + pi + tan^-1(\frac {2+3}{1-2 3}) because [tan^-1(A) + tan^-1(B) =pi + tan^-1(\frac {A+B}{1 - A B)}] when A*B > 1 \Rightarrow pi/4 + pi + tan^-1(-1) \Rightarrow pi/4 + pi + (-pi/4) \Rightarrow pi/4 + pi - pi/4 \Rightarrow pi.