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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.

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Note by Calvin Lin
3 years, 4 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!! Abhinav Raichur · 3 years, 2 months ago

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