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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{align} & \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{align}

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
3 years, 6 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!!

- 3 years, 5 months ago

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