Hardest Inverse and Binomial questions? (1 Viewer)
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Paradoxica
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Q1: The 3 Pythagorean triples are: {3,4,5}, {5,12,13} and {16,63,65} = {3.2,13,12.6}} \equiv 2\tan^{-1}{t} \\ $Hence, or otherwise, show/prove $\tan^{-1}{\frac{1}{3}} + \tan^{-1}{\frac{2}{3}} + \tan^{-1}{\frac{7}{9}} = \frac{\pi}{2})
You can draw the 3 right-angled triangles associate with these triples: {3,4,5}, {5,12,13} and {3.2,12.6,13}
You draw the 1st triangle with side 4 horizontal. You then erect the 2nd right-angled triangle with the 5 side resting on the hypotenuse of the 1st. You finally allign the 13 (hypotenuse side) of the 3rd triangle withe the 13 (hypotenuse) side of the 2nd triangle. Then there you have it: all the 3 adjacent acute angles:
forming a straight angle.
Of course a simple diagram would have made this easier to see; but I just can't do it. Some kind soul can easily do this and display the diagram.
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Q2
The 1st part is the usual t-formulae
Now applying this result (omitting much detail because of LaTeX tedium):
^2} \right )} = sin^{-1}\frac{3}{5}\\ \\ 2tan^{-1}\frac {2}{3} = sin^{-1} \frac {12}{13}\\ \\ 2tan^{-1}\frac {7}{9} = sin^{-1}\frac {63}{65}\\ \\ \therefore 2(tan^{-1}\frac {1}{3}+tan^{-1}\frac{2}{3}+tan^{-1}\frac {7}{9}) \\ \\ = sin^{-1}\frac{3}{5} + sin^{-1}\frac {12}{13} + sin^{-1}\frac {63}{65}= \pi)
by virtue of Q1 result.
Hence the result.
Alternatively, by brute force:
By repeated application of the formula:
 = \frac {tan \alpha + tan \beta}{1 - tan\alpha \tan \beta}\\ \\ $you will end up with $ tan($sum of the 3 inverse tangents $) \\ \\ = \frac {\frac {1}{3} + 3}{1 - \frac{1}{3} \times 3} = \frac {\frac {10}{3}}{1 - 1} = \infty)
So that angle must be pi/2
The 1st part is the usual t-formulae
Now applying this result (omitting much detail because of LaTeX tedium):
by virtue of Q1 result.
Hence the result.
Alternatively, by brute force:
By repeated application of the formula:
So that angle must be pi/2
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