$$DR_t= \frac{B_t}{P_t}-\frac{B_{t-1}}{P_{t-1}}\tag{1}$$
$$DR_t=\frac{NB_t}{P_t}- \pi_t\frac{B_{t-1}}{P_t}\tag{2}$$
Knowing that $\pi_t=\frac{P_t}{P_{t-1}}-1$ and $NB_t=B_t - B_{t-1}$, demonstrate that $(1)$ is equal to $(2)$.
So what I have done is substituted $\pi_t$ and $NB_t$ into $(2)$ to get:
$$DR_t= \frac{B_t - B_{t-1}}{P_t} - \left(\frac{Pt}{P_{t-1}} -1\right) \frac{B_{t-1}}{P_t}$$
After that I made the following equation:
$$\frac{B_t - B_{t-1}}{P_t} - \left(\frac{Pt}{P_{t-1}} -1 \right)\frac{B_{t-1}}{P_t} = \frac{B_t}{P_t} - \frac{B_{t-1}}{P_{t-1}}$$
Now I have been trying to solve this equation in a variety of ways, since the first thing I tried (moving everything to one side and try to factor it all out) did not work. But without success.
Normally, I would go to my professor for help, but since we have to hand this in first thing after the weekend and got the assignment last Friday, I do not have time to do it.
A hint or just some recommended reading material in order to finish the assignment would suffice as well.
Thanks :)
I don´t know what went wrong at your transformations. My steps are the following:
$$\frac{B_t - B_{t-1}}{P_t} - \left(\frac{P_t}{P_{t-1}} -1 \right)\frac{B_{t-1}}{P_t} $$
The first fraction can be splitted in two fractions:
$$\frac{B_t }{P_t}-\frac{B_{t-1}}{P_t} - \left(\frac{P_t}{P_{t-1}} -1 \right)\frac{B_{t-1}}{P_t} $$
Multiplying out the brackets
$$\frac{B_t }{P_t}\color{blue}{-\frac{B_{t-1}}{P_t}} - \frac{P_t}{P_{t-1}}\cdot \frac{B_{t-1}}{P_t} \color{blue}{+\frac{B_{t-1}}{P_t}}$$
The sum of the blue terms are equal to zero. Finally cancel out.