# Summation inside tan(x)

Given $$a_1 +a_2 + a_3+...+a_n= \theta$$ degrees. Where $$tan(a_k) = \frac{n}{n^2 + k(k-1)}$$. Find $$tan(\theta)$$ in terms of "n".

I tried using the formula tan(a+b+c+d+...) = $$\frac{S_1-S_3-S_5-...}{1-S_2-S_4-...}$$, where $$S_n$$ denotes summation of tan(x), taken 'n' at a time. But it was proving to be quite difficult.

Can anyone help me with this problem? Is there some sort of visual solution to this?

In fact, rewriting $$\tan(a_k)$$ under the form :

$$\tan(a_k)=\frac{\tfrac{k}{n}-\tfrac{k-1}{n}}{1+\tfrac{k}{n}\tfrac{k-1}{n}}$$

we recognize in the RHS the formula of

$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$

Therefore, setting :

$$\alpha_k=\operatorname{atan}(\tfrac{k}{n}),$$

the given sum becomes :

$$\theta = (\alpha_n-\alpha_{n-1})+(\alpha_{n-1}-\alpha_{n-2})+\cdots+(\alpha_{1}-\alpha_{0})$$

which is a telescopic sum.

Therefore :

$$\theta=\alpha_n= \operatorname{atan}(\tfrac{n}{n})=\frac{\pi}{4},$$

whatever the value of $$n.$$

• Thanks! I did not think the solution would be so simple and also it is independent of n. Nov 28, 2023 at 5:33