Ideal low pass filter output at given sampling frequency
That $\frac{1}{T_s}$ factor in your sampled signal expression is $\frac{20}{\pi}$.

$$\omega_s = 40 \ rad/sec$$
$$2\pi f_s = 40$$
$$f_s = \frac{40}{2\pi} = \frac{20}{\pi}$$

The sampled signal is basically given by :
$$X_{sampled}(f) = f_s \sum^{\infty}_{k=-\infty} X(f - kf_s)$$
Meaning : Sampling a signal with impulse train at sampling frequency $f_s$ gives you, images of $X(f)$ scaled by $f_s$ and centered at $kf_s$.

You can check this answer to understand why : Sampling with Impulse Train explained pictorially

The original signal is at frequency $\frac{15}{\pi}$ and being sampled at sampling frequency $\frac{20}{\pi}$. And hence the frequency components in the sampled signal will be : $\pm k\frac{20}{\pi} \pm \frac{15}{\pi}$.

Notice that for $k=0$, you get the original signal and for $k=\pm 1$, you have an alias term $\cos{(2\pi (\frac{20}{\pi} - \frac{15}{\pi})t)} = cos(10t)$ which will get inside the lowpass filter cutoff frequency.

And, that will give you the final result as :
$$f_s (\cos{(2\pi \frac{15}{\pi}t)} + \cos{(2\pi \frac{5}{\pi}t)})$$ 
$$ = \frac{20}{\pi} (\cos(30t) + \cos(10t))$$