Trigonometry - simplifying a given equation

Question

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

Answer I'm getting : 0

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)

Answer 1

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 $8.(\frac{1}{\sqrt 5 -1}-\frac{1}{\sqrt 5 +1})=4$

Answer 2

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)