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I have solved half of the problem by taking $\tan(A+B) = \tan(\pi -C)$. But I am stuck in the middle. So how to prove the statement ?

Leucippus
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Suman Tiwari
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    Sorry, what is your question. Something seems to have been lost in translation. And why don't you show us what you have done. – Doug M Sep 15 '18 at 01:38
  • See https://math.stackexchange.com/questions/446402/if-alpha-beta-dfrac-pi4-prove-that-1-tan-alpha1-tan-beta OR https://math.stackexchange.com/questions/1716859/how-to-calculate-left-1-tan-5-circ-right-left-1-tan-10-circ-right-lef OR https://math.stackexchange.com/questions/745929/calculate-for-1-tan-20-circ1-tan-25-circ-help-me-with-my-works – lab bhattacharjee Sep 15 '18 at 10:05
  • Also : https://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt – lab bhattacharjee Sep 15 '18 at 10:05

1 Answers1

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It is certainly true that

$1+\tan 10°=\dfrac{\sin 10° + \cos 10°}{\cos 10°}=(\sqrt{2})(\dfrac{\sin (10°+45°)}{\cos 10°})$

where $\sin (10°+45°)=(\sin 10°/\sqrt{2})+(\cos 10°/\sqrt{2})$ from the formula for the sine of a sum. Then, continuing:

$1+\tan 10°=(\sqrt{2})(\dfrac{\sin (10°+45°)}{\cos 10°})=(\sqrt{2})(\dfrac{\cos 35°}{\cos 10°})$

using $\sin (10°+45°)=\cos (90°-10°-45°)=\cos 35°$. Do the same with arguments of $20°, 25°, 35°$ in place of $10°$ and multiply the four resulting fractions together; all the trig functions cancel out of the product and you have just $(\sqrt{2})^4=4$.

The business with $A, B, C$, however, has me completely stumped. It does not enter the above equality at all!

Oscar Lanzi
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