2015年9月9日 · Evaluate $$\tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ$$ using trigonometric ratios of angles $0^o,\ 30^o,\ 45^o,\ 60^o,\ 90^o$ I played with this problem for a while, but I still can't figure out how to solve it.

2020年8月19日 · The numerator is -tan20°+tan40°-tan80° this means that +tan40° should be an excess value because tan40° was approximated by excess two times while tan80° was approximated by defect slightly, then I add tan20° that is a value approximated by defect, so the numerator should be a too-high value.

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The value of tan40°+tan20°+√3tan20°tan40° is

2015年7月23日 · The equality I'm trying to prove looks like that: $$ \tan40° + \sqrt 3 =4 \sin40° $$ My guess is that $\sqrt3$ can be rewritten as $\tan60°$ and I can use proved in previous exercise formula $$\tan3 \alpha = \frac{(3 - \tan^2\alpha)\tan\alpha}{1 - 3\tan\alpha}$$ But after dealing with it for an hour I feel misled.

Prove that tan 50° tan 40° = 2 tan 10°

2019年5月29日 · $$\tan20\tan40\tan60\tan80$$ I tried to bring $\tan20$, $\tan40$ in $\tan(a+b)$ form but couldn't get the answer. Using calculator I got it has $3$. Can anybody please tell me how to solve this?

2016年2月4日 · I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ as $\tan 20^\circ \tan40^\circ \tan120^\circ$. But I really can't get to the end of it . Please help.

Notice that for $\theta = 20, 40,$ and $80$ degrees you have $\tan^2(3\theta) = 3$. The tangent triple angle formula, which you can get from the tangent angle addition formula, says that $$\tan(3\theta) = {3\tan(\theta) - \tan^3(\theta) \over 1 - 3 \tan^2(\theta)}$$ So the equation $\tan^2(3\theta) = 3$ can be expressed as $$(3\tan(\theta) - \tan^3(\theta))^2 = 3(1 - 3 …

(Prove that) tan40 + 2 tan10, = tan50