# 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)

• 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. Commented Jul 10, 2014 at 12:42
• $0\cdot\infty$ is undefined. Not saying that's your problem, just saying... Commented Jul 10, 2014 at 12:43
• @J.Finnegan I didn't use a calculator? Commented Jul 10, 2014 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. Commented Jul 10, 2014 at 12:45
• According to W|A this is $4$, not $0$... Commented Jul 10, 2014 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 $8.(\frac{1}{\sqrt 5 -1}-\frac{1}{\sqrt 5 +1})=4$

• 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. Commented Jul 10, 2014 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
Commented Jul 10, 2014 at 12:51
– puru
Commented Jul 10, 2014 at 12:52
• I see. Well the answer is supposed to be four as has been said in my comments. Is your answer matching? Commented Jul 10, 2014 at 12:53
• @Gummybears Yeah, it is coming out to be 4.
– puru
Commented Jul 10, 2014 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)

• 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. Commented Jul 10, 2014 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". Commented Jul 10, 2014 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
Commented Jul 10, 2014 at 13:03
• Yes, that was meant to be my point, thanks @puru Commented Jul 10, 2014 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." Commented Jul 10, 2014 at 13:08