How to evaluate $\tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ$ using trigonometric ratios of angles $0^o, 30^o, 45^o ,60^o, 90^o$

Question

The question is

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.

All I could determine was,

$$\tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ$$

$$=({\tan60^\circ})(1-\tan20^\circ\tan40^\circ) + \sqrt3\tan20^\circ\tan40^\circ$$

Now I don't know how to proceed.

what happens if you just open the bracket and multiply? It is a straight forward cancellation with just $\tan 60^\circ$ remaining. Do you see it? — MonK, Sep 09 '15 at 16:01
It was so simple and I didn't notice it. But my evaluation was wrong at first and the question was edited later as per the suggestion of authors. — H G Sur, Sep 09 '15 at 16:14

score 2 · Accepted Answer · answered Sep 09 '15 at 16:01

2

Let me try.

One has $$\tan(\alpha+\beta) = \frac{\tan \alpha + \tan\beta}{1-\tan\alpha\tan\beta}$$

One then has $$\sqrt{3} = \tan 60^\circ = \tan(20^\circ+40^\circ) = \frac{\tan 20^\circ + \tan 40^\circ}{1-\tan 20^\circ \tan 40^\circ}$$

$$\tan 20^\circ + \tan 40^\circ + \sqrt{3}\tan 20^\circ \tan 40^\circ = \sqrt{3}.$$

answered Sep 09 '15 at 16:01

GAVD

I have difficulties in understanding the last step :( – H G Sur Sep 09 '15 at 16:06
1

Uhm, close your eyes, relax. After that, you can see why :) – GAVD Sep 09 '15 at 16:08
Yes, you cross-multiplied it ? :) – H G Sur Sep 09 '15 at 16:10

score 1 · Answer 2 · answered Sep 09 '15 at 15:43

1

Hint: $20^{\circ} + 40^{\circ}=60^{\circ}$

Taking $\tan$ on both sides: $$\tan(20^{\circ} + 40^{\circ}) =\tan(60^{\circ})$$ Now just expand.

answered Sep 09 '15 at 15:43

Is the answer $\sqrt3$? – H G Sur Sep 09 '15 at 15:47
1

Yes it is $\sqrt{3}$ – Sep 09 '15 at 15:48
Thank you sir, but is there a way to evaluate it from the question itself? – H G Sur Sep 09 '15 at 15:49
1

@HGSur the right side of the expression is incorrect because tan(60) is not tan20 + tan40 – user258250 Sep 09 '15 at 15:51
I don't understand what you said, Sir. – H G Sur Sep 09 '15 at 15:54
I found out my mistake – H G Sur Sep 09 '15 at 15:56
@HGSur in your equation you can cancel 1 - tan20tan40 on the right side but you see that tan20+tan40+sqrt(3)tan20tan40 = tan60 +sqrt(3)tan20tan40 which is incorrect. – user258250 Sep 09 '15 at 15:57
@user258250 I've edited the question, is it now correct? – H G Sur Sep 09 '15 at 15:59

2 Answers2