1. 2 hard trig problems

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

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

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

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

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

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

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

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

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

$=$ ... etc

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

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

Can you explain what identity you used ??

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

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