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.

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#Algebra #Proofs #Olympiad

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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 - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$