# Trigonometric proof that sum of positive numbers=0

The sum of tangents trigonometric identity gives us that

$\tan ( \alpha + \beta) = \frac{ \tan \alpha + \tan \beta} { 1 - \tan \alpha \tan \beta} .$

By letting $\alpha = \tan^{-1} x$ and $\beta = \tan^{-1} y$, the equivalent trigonometric identity on $\tan^{-1}$ is

$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y} { 1-xy} \right).$

Let's use this identity to calculate $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$. We first have

$\tan^{-1} 1 + \tan^{-1} 2 = \tan^{-1} \left( \frac { 1 + 2 } { 1 - 1 \times 2} \right)= \tan^{-1} (-3).$

As such, this gives

\begin{aligned} & \tan^{-1} 1+ \tan^{-1} 2 + \tan^{-1} 3 \\ = & \tan^{-1} (-3) + \tan^{-1} 3\\ = & \tan^{-1} \left( \frac{-3 + 3} { 1 - (-3)\times 3 } \right) \\ = & \tan^{-1} 0 \\ = & 0 \\ \end{aligned}

By considering the right triangles, we get that $\tan^{-1} 1 > 0$, $\tan^{-1} 2 > 0$ and $\tan^{-1} 3 > 0$. Hence, the sum of 3 positive terms is 0.

What went wrong?

Note by Calvin Lin
7 years, 2 months ago

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I think you missed the condition there ...... Mod(xy) should be less than one to use the formula .... Or els the RHS becomes negative where as LHS remains positive!!

- 7 years, 1 month ago