sequences and series - 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.

Answer

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.