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 $n.$