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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?

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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 $n.$

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

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