Simplify $\sum_{k = 1}^n \tan(k) \tan(k - 1)$ by first proving $\tan(k)\tan(k - 1) = \frac{\tan(k) - \tan(k - 1)}{\tan(1)} - 1$ I have the following problem:

Use the formula
$$\tan(A - B) = \dfrac{\tan(A) - \tan(B)}{1 + \tan(A) \tan(B)}$$
to prove that
$$\tan(k)\tan(k - 1) = \dfrac{\tan(k) - \tan(k - 1)}{\tan(1)} - 1$$
Hence simplify
$$\sum_{k = 1}^n \tan(k)\tan(k - 1)$$

Since we have that
$$\tan(A - B) = \dfrac{\tan(A) - \tan(B)}{1 + \tan(A) \tan(B)},$$
I then deduced that
$$\tan(k)\tan(k - 1) = \dfrac{\tan(k)[\tan(k) - \tan(-1)]}{1 + \tan(k)\tan(-1)}$$
But I'm not sure how to proceed from here. And I don't have any solutions to refer to.
I would greatly appreciate it if people could please take the time to clarify this.
 A: HINT:
Note that $$\tan1=\tan(k-(k-1))=\frac{\tan k-\tan(k-1)}{1+\tan k\tan(k-1)}$$ from which the result follows.
The summation part is easy as the numerator is telescoping.
A: Let $a:=\tan A,\,b:=\tan B,\,c:=\tan(A-B)=\dfrac{a-b}{1+ab}$ so $ab=\dfrac{a-b}{c}-1$. Hence $\tan k\tan (k-1)=\dfrac{\tan k-\tan (k-1)}{\tan 1}-1$, and your sum is $\dfrac{\tan n}{\tan 1}-n$.
A: $$\tan(k-(k-1))=\frac{\tan k-\tan(k-1)}{1+\tan k \tan (k-1)}$$
$$1+\tan k \tan (k-1)=\frac{\tan k-\tan(k-1)}{\tan 1}$$
$$\tan k \tan (k-1)=\frac{\tan k-\tan(k-1)}{\tan 1}-1$$
A: First, we have this
$$tan(A−B)= \frac{tan(A)−tan(B)}{1+tan(A)tan(B)}$$
Then, we can multiply both sides by $1 + tan(A)tan(B)$
$$tan(A−B)(1 + tan(A)tan(B))= \frac{(tan(A)−tan(B))(1 + tan(A)tan(B))}{1+tan(A)tan(B)}$$
Now we can simplify in the right hand side the expression $1 + tan(A)tan(B)$ and then, divide both sides by $tan(A-B)$
$$\frac{tan(A−B)(1 + tan(A)tan(B))}{tan(A - B)}= \frac{tan(A)−tan(B)}{tan(A - B)}$$
Now, simplify $tan(A - B)$
$$1 + tan(A)tan(B)= \frac{tan(A)−tan(B)}{tan(A - B)}$$
Now, let's substract 1 in both sides
$$tan(A)tan(B)= \frac{tan(A)−tan(B)}{tan(A - B)} - 1$$
Now, Let $A = k$ and $B = (k - 1)$
$$tan(k)tan(k - 1)= \frac{tan(k)−tan(k - 1)}{tan(1)} - 1$$
Note: $tan(A - B)$ can't by equal to zero to divide
