# Thread: 2 hard trig problems

1. ## 2 hard trig problems

I have tried to solve these through every way i know but i get stuck halfway.

$\displaystyle tan^{2}10^{\circ}+tan^{2}50^{\circ}+tan^{2}70^{\ci rc}=9$

$\displaystyle tan^{2}36^{\circ}\times tan^{2}72^{\circ}=5$

2. Hello ldx2
Originally Posted by ldx2
I have tried to solve these through every way i know but i get stuck halfway.

$\displaystyle tan^{2}10^{\circ}+tan^{2}50^{\circ}+tan^{2}70^{\ci rc}=9$

$\displaystyle tan^{2}36^{\circ}\times tan^{2}72^{\circ}=5$
I haven't got a solution for part 1 yet, but here's an answer to part 2, based on the fact that $\displaystyle \sin18^o = \frac{\sqrt5-1}{4}$ (which we proved in an earlier post).

First note that
$\displaystyle \sin18^o = \frac{\sqrt5-1}{4}$
$\displaystyle \Rightarrow \sin^218^o = \frac{6-2\sqrt5}{16}=\frac{3-\sqrt5}{8}$
Then, noting that $\displaystyle \tan72^o = \cot18^o = \frac{1}{\tan18^o}$:
$\displaystyle \tan36^o\tan72^o = \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}$
$\displaystyle =\frac{2}{1-\dfrac{\sin^218^o}{\cos^218^o}}$

$\displaystyle =\frac{2(1-\sin^218^o)}{1-2\sin^218^o}$

$\displaystyle = \frac{2\left(1-\dfrac{3-\sqrt5}{8}\right)}{1-\dfrac{3-\sqrt5}{4}}$

$\displaystyle =$ ... etc

$\displaystyle = \sqrt5$ (Can you supply the missing lines?)
Hence $\displaystyle \tan^236^o\tan^272^o=5$

3. I understood all but this part :
$\displaystyle \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}$

Can you explain what identity you used ??

4. Hello ldx2
Originally Posted by ldx2
I understood all but this part :
$\displaystyle \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}$

Can you explain what identity you used ??
$\displaystyle \tan2A = \frac{2\tan A}{1-\tan^2A}$. So $\displaystyle \tan36^o =$ ?