1. **State the problem:** We are given the function $f(t) = t$ defined on the interval $0 < t < \pi$. We want to analyze or solve for properties related to this function. 2. **Formula and rules:** Since the problem does not specify what to solve, a common task is to find the Fourier series representation of $f(t)$ on $0 < t < \pi$ or to analyze its behavior. 3. **Fourier sine series for $f(t)$ on $(0, \pi)$:** The sine series coefficients are given by $$b_n = \frac{2}{\pi} \int_0^{\pi} t \sin(nt) \, dt$$ 4. **Calculate $b_n$:** Use integration by parts: Let $u = t$, $dv = \sin(nt) dt$, then $du = dt$, $v = -\frac{\cos(nt)}{n}$. $$b_n = \frac{2}{\pi} \left[-\frac{t \cos(nt)}{n} \Big|_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos(nt) dt \right]$$ 5. **Evaluate boundary term:** $$-\frac{\pi \cos(n\pi)}{n} + 0 = -\frac{\pi (-1)^n}{n}$$ 6. **Evaluate integral:** $$\int_0^{\pi} \cos(nt) dt = \frac{\sin(n\pi)}{n} - \frac{\sin(0)}{n} = 0$$ 7. **Simplify $b_n$:** $$b_n = \frac{2}{\pi} \left(-\frac{\pi (-1)^n}{n} + 0 \right) = \frac{2}{\pi} \left(-\frac{\pi (-1)^n}{n} \right) = -\frac{2 (-1)^n}{n}$$ 8. **Final Fourier sine series:** $$f(t) = \sum_{n=1}^\infty b_n \sin(nt) = -2 \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(nt)$$ 9. **Interpretation:** This series represents the function $f(t) = t$ on the interval $(0, \pi)$ as a sum of sine functions. **Answer:** $$f(t) = -2 \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(nt)$$