How can I prove that $T_\pi = 0$ at Brewster's angle? So, I've been trying to prove this with pure trigonometry just for fun, without using the fact that $R + T = 1$, but no success. Here's my last try, using a combination of both facts that the angles are complementary and Snell's Law:
\begin{align*}
t_\pi &= \frac{2n_i\cos(\theta_i)}{n_i\cos(\theta_t) + n_t\cos(\theta_i)}\\
&= \frac{2n_i\cos(\theta_i)}{n_t\cos(\theta_i) + n_t\cos(\theta_i)}\\
&= \frac{2n_i\cos(\theta_i)}{2n_t\cos(\theta_i)}\\
&= \frac{n_i}{n_t}
\end{align*}
Now, either I suppose $n_t = n_i$ (which is absurd) or I'm wrong at something.
 A: $t$ is the transmission coefficient.
$T$ is the Transmittance which is equal to 1 at brewsters angle since $R$, Reflectance goes to zero.
Also,
$$T = n\frac{\cos\theta_t}{\cos\theta_i}|t|^2$$
A: Sumit Gupta essentially answered the question. However, I think it's not wrong to elaborate a bit further. Your calculation is entirely correct. Indeed,
$$ t_p = \frac{n_i}{n_t}. $$
How does that relate to the well-known fact $T_\pi = 1$ at Brewster's angle? Transmittance is defined as the power transported by the transmitted field divided by the power of the incoming field. The power of a light field, on the other hand, is the integral of its intensity over its section area. Let's take a look at those quantities.
Intensity
The intensity of an electromagnetic wave is the magnitude of the Poynting vector
$$ \mathbf{S} = \frac{1}{2} \mathbf{E} \times \mathbf{H^*}$$
For a plane wave in a non-magnetic medium, this is just
$$|\mathbf{S}| = \frac{1}{2} n |E_0|^2$$
(c.f. J.D. Jackson Classical Electrodynamics, Chapter 7). Note that the intensity depends on the index of refraction. Intuitively, this makes sense because the flow of energy is correlated with the propagation of the EM wave.
Since the indices of refraction are different for the transmitted and incoming waves, the transmittance can't just be the ratio of their amplitudes but has to take account of the refractive indices as well. Therefore, we would write
$$T_\pi \propto \frac{n_t}{n_i} |t_\pi|^2.$$
Section area
One could replace the $\propto$ above by an $=$ if the section areas of the two waves were the same. However, as the incoming and transmitted beams have the same projection on the boundary, their section areas will generally differ. It is a simple geometrical exercise to show that the section area is proprtional to
$$\cos(\theta)$$
if $\theta$ is the angle of incidence. Therefore, the transmittance is
$$T_\pi = \frac{n_t \cos(\theta_t)}{n_i \cos(\theta_i)} |t_\pi|^2.$$
Conclusion
Putting everything together, we have
$$T_\pi = \frac{n_t \cos(\theta_t)}{n_i \cos(\theta_i)} \left| \frac{n_i}{n_t} \right|^2 = \frac{n_i \cos(\theta_t)}{n_t \cos(\theta_i)} = \frac{n_i}{n_t} \tan(\theta_i) = 1$$
since $\theta_i = \arctan(n_t / n_i)$ at Brewster's angle.
