Gabriela Duarte

2022-01-27

Simplify $\sum _{k=1}^{n}\mathrm{tan}\left(k\right)\mathrm{tan}\left(k-1\right)$ by first proving $\mathrm{tan}\left(k\right)\mathrm{tan}\left(k-1\right)=\frac{\mathrm{tan}\left(k\right)-\mathrm{tan}\left(k-1\right)}{\mathrm{tan}\left(1\right)}-1$

goleuedigdp

Expert

HINT:
Note that
$\mathrm{tan}1=\mathrm{tan}\left(k-\left(k-1\right)\right)=\frac{\mathrm{tan}k-\mathrm{tan}\left(k-1\right)}{1+\mathrm{tan}k\mathrm{tan}\left(k-1\right)}$
from which the result follows.
The summation part is easy as the numerator is telescoping.
$\sum _{k=1}^{n}\left[\frac{\mathrm{tan}k-\mathrm{tan}\left(k-1\right)}{\mathrm{tan}1}-1\right]=\frac{\mathrm{tan}n-\mathrm{tan}\left(n-1\right)+\mathrm{tan}\left(n-1\right)-\mathrm{tan}\left(n-2\right)+\cdots +\mathrm{tan}1-\mathrm{tan}0}{\mathrm{tan}1}-n=\frac{\mathrm{tan}n}{\mathrm{tan}1}-n$

trasahed

Expert

Let so $ab=\frac{a-b}{c}-1$. Hence $\mathrm{tan}k\mathrm{tan}\left(k-1\right)=\frac{\mathrm{tan}k-\mathrm{tan}\left(k-1\right)}{\mathrm{tan}1}-1$ and your sum is $\frac{\mathrm{tan}n}{\mathrm{tan}1}-n$