Is it permitted to use $\tan 90^\circ$ to prove an equation/identity even if it is not defined? defined? Domain of tangent function limits to all real numbers except odd integral multiples of $\frac{\pi}{2}$.
I noticed when proving an equality that using $\tan 90^\circ$ yielded the exact same result. Here what I did:

For a triangle $\mathrm{ABC}$, prove that $$\tan \frac{A}{2}\tan\frac{ B}{2 }+\tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2}=1$$

Since $\mathrm{A+B+C = \pi}$
$$\mathrm{\frac{A+B+C}{2} = \frac{\pi}{2}}$$
I applied identity for sum of angles
$$\tan{(A/2 + B/2 + C/2)}=\frac{\tan \frac{A}{2} +\tan \frac{B}{2} +\tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{1-\tan \frac{A}{2}\tan\frac{ B}{2 }-\tan \frac{B}{2} \tan \frac{C}{2} - \tan \frac{C}{2}\tan \frac{A}{2}}$$
Then I reciprocated both side to get L.H.S. to be equal to 0 and I was able to prove the given equation.
Why did I arrive at the right conclusion by using the value out of domain for tangent function?
 A: Just like there is an identity for the tangent of a sum of three angles, there is an identity for the cotangent of a sum of three angles.
Not surprisingly, you can write a version of it just by taking the reciprocal of the formula for the tangent of the sum:
$$
\cot\left(\alpha + \beta + \gamma\right)
= \frac{1 -\tan \alpha \tan\beta - \tan\beta \tan\gamma - \tan\gamma \tan\alpha}
       {\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha \tan\beta \tan\gamma}.
$$
Plug in $\alpha = \frac A2,$ $\beta = \frac B2,$ $\gamma = \frac C2$:
$$
\cot\left(\frac A2 + \frac B2 + \frac C2\right)
= \frac{1 - \tan\frac A2 \tan\frac B2 - \tan\frac B2 \tan\frac C2 - \tan\frac C2 \tan\frac A2}
{\tan\frac A2 + \tan\frac B2 + \tan\frac C2 - \tan\frac A2 \tan\frac B2 \tan\frac C2}.
$$
Now just use the fact that $\cot\left(\frac\pi2\right) = 0.$
Intuitively, this seems to be what you were trying to do.
The main difference in the methods is that $\cot\left(\frac\pi2\right) = 0$
is a true equation with well-defined quantities on each side, wherease
$1/\tan\left(\frac\pi2\right) \stackrel?= 0$ is not.
