# Trigonometry - simplifying a given equation [duplicate]

Question: $$\tan 9 - \tan 27 - \tan 63 + \tan 81$$

What I did: Well I clubbed together $\tan 9$ and $\tan 81$ and $\tan 27$ and $\tan 63$ (took out negative as common). Then using the identity for $\tan (A+B)$, I rearranged to formula to get what $\tan A + \tan B$ is. With that I'm getting zero multiplied by $\tan 90$. Since anything multiplied by zero, even infinity, is zero, I guess it should be zero.

I'm pretty sure my logic fails me somewhere, please tell me where (probably in the infinity and zero multiplication part)

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Firstly, it would seem as if the argument of the tangent functions is in degrees. Calculators often take the arguments in terms of radians, so this may be where you went wrong. – user138999 Jul 10 '14 at 12:42
$0\cdot\infty$ is undefined. Not saying that's your problem, just saying... – gebruiker Jul 10 '14 at 12:43
@J.Finnegan I didn't use a calculator? – Gummy bears Jul 10 '14 at 12:44
@gebruiker Well I was pretty sure that's where my problem was. Still, then I have nothing else to offer to the question :/Then I need help solving the question. – Gummy bears Jul 10 '14 at 12:45
According to W|A this is $4$, not $0$... – draks ... Jul 10 '14 at 12:46

HINT:

$\tan 9+\tan 81=\tan 9+\frac{1}{\tan 9}=\frac{\sec^2 9}{\tan 9}=\frac{1}{\sin 9 \cos 9}=\frac{2}{\sin 18}$

Similarly, $\tan 27+\tan 63=\frac{2}{\sin 54}$

Here you can find the derivation of value of $\sin 18$ and $\sin 54$(or)$\cos 36$. Plug in the values and get the answer.Hope this helps. Correct me if I'm wrong!

So $\sin 18=\frac {\sqrt 5 -1}{4}$ and $\sin 54= \frac {\sqrt 5 + 1}{4}$. So we get $8.(\frac{1}{\sqrt 5 -1}-\frac{1}{\sqrt 5 +1})=4$

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Well the answer might be correct, but the fact is that it is supposed to be done without a calculator. So there has to be a way to simplify and find the value of this question easily, without calculator. – Gummy bears Jul 10 '14 at 12:49
@Gummybears we are not using calculators at all. The pdf shows the values calculated by using identities. Moreover, the irrational part cancels out once you rationalize. There's no need of calculator. – puru Jul 10 '14 at 12:51
I'll add the details to the answer soon. – puru Jul 10 '14 at 12:52
I see. Well the answer is supposed to be four as has been said in my comments. Is your answer matching? – Gummy bears Jul 10 '14 at 12:53
@Gummybears Yeah, it is coming out to be 4. – puru Jul 10 '14 at 12:54

You have $$\tan (9+81)=\frac{\tan 9 +\tan 81}{1-\tan 9\tan 81}\\\infty=\frac{\tan9+\tan81}{0}$$ So it is impossible to determine the value of $\tan9+\tan81$ using this formula." . (thanks Theophile)

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The formula you give works precisely in the case when the two things you are adding don't add up to (working in radians here) $\pi/2 + n\pi$ for some integer $n$ - otherwise $\tan(\pi/10)$ and $\tan(9\pi/10)$ are perfectly well defined, and so is the result by adding them. – Andrew D Jul 10 '14 at 13:00
This is not a valid answer. Tan9, tan27, tan 63 and tan81 are all real and unique numbers. When you add and substract them, you still have a real and unique number, and not "any number". – Martigan Jul 10 '14 at 13:01
I think Michael is trying to point out the mistake in the Gummy bears' reasoning. I think he wants to say that the number could have been anything and not necessarily 0. – puru Jul 10 '14 at 13:03
Yes, that was meant to be my point, thanks @puru – Michael Jul 10 '14 at 13:04
I think a clearer way to put the last sentence would be: "So it is impossible to determine the value of $\tan 9^\circ + \tan 81^\circ$ using this formula." – Théophile Jul 10 '14 at 13:08