A trig identity

Let $t = \tan \theta$ , then using the half-angle identity of cosine, we have:

$\begin{aligned} \cos (2\theta) & =\frac{1-\tan^2 \theta}{1+\tan^2 \theta} \\ \Rightarrow \frac{1-t^2}{1+t^2} & = \frac{1+\sqrt{5}}{4} \\ 4 - 4t^2 & = 1+\sqrt{5} + (1+\sqrt{5})t^2 \\ (5+\sqrt{5})t^2 & = 3 - \sqrt{5} \\ \Rightarrow t^2 & = \frac{3 - \sqrt{5}}{5+\sqrt{5}} = \frac{(3 - \sqrt{5})(5-\sqrt{5})}{25-5} = \frac{20-8\sqrt{5}}{20} = \frac{\sqrt{5} - 2}{\sqrt{5}} \end{aligned}$

$\begin{aligned} \tan^2 (3 \theta) \tan^2 \theta & = \left(\frac{t(3-t^2)}{1-3t^2}\right)^2t^2 \\ & = \frac{t^4(3-t^2)^2}{(1-3t^2)^2} \\ & = \left(\frac{(\sqrt{5}-2)(2\sqrt{5}+2)}{\sqrt{5}(6-2\sqrt{5})}\right)^2 \\ & = \left(\frac{(\sqrt{5}-2)(\sqrt{5}+1)}{\sqrt{5}(3-\sqrt{5})}\right)^2 \\ & = \left(\frac{3-\sqrt{5}}{\sqrt{5}(3-\sqrt{5})}\right)^2 \\ & = \frac{1}{5} \end{aligned}$

$\Rightarrow m + n = 1 + 5 = \boxed{6}$

Recall: $\cos{\left(\theta_1+\theta_2\right)}= \cos{\theta_1}\cos{\theta_2} - \sin{\theta_1}\sin{\theta_2}$ and $\cos{\left(\theta_1-\theta_2\right)} = \cos{\theta_1}\cos{\theta_2} + \sin{\theta_1}\sin{\theta_2}$ Thus we have: $\tan^2{(3\theta)}\tan^2{\left(\theta\right)} = \left(\frac{\sin{3\theta}\sin{\theta}}{\cos{3\theta}\cos{\theta}}\right)^2$

$\sin{3\theta}\sin{\theta} = \sin{3\theta}\sin{\theta} - \sin{3\theta}\sin{\theta} + \cos{3\theta}\cos{\theta} -\cos{3\theta}\cos{\theta} + \sin{3\theta}\sin{\theta}$

$2\sin{3\theta}\sin{\theta} = \sin{3\theta}\sin{\theta} - \sin{3\theta}\sin{\theta} + \cos{3\theta}\cos{\theta} - \cos{3\theta}\cos{\theta}$

$\sin{3\theta}\sin{\theta} = \frac{ \left(\cos{3\theta}\cos{\theta}+\sin{3\theta}\sin{\theta}\right) -\left(\cos{3\theta}\cos{\theta}+\sin{3\theta}\sin{\theta}\right)}{2}$

$\sin{3\theta}\sin{\theta} = \frac{1}{2} \left( \cos{\left(3\theta - \theta)\right)} - \cos{\left(3\theta+\theta\right)}\right) = \frac{1}{2}\left(\cos{2\theta} - \cos{4\theta}\right)$

We can transform the cosine terms in a similar fashion and we get:

$\cos{3\theta}\cos{\theta}=\frac{1}{2} \left(\cos{2\theta}+\cos{4\theta}\right)$

Thus our combined expression looks as follows:

$\tan^2{(3\theta)}\tan^2{\left(\theta\right)} = \left(\frac{\cos{2\theta} - \cos{4\theta}}{\cos{2\theta} + \cos{4\theta}}\right)^2$

Recall: $\\ \cos{4\theta} = \cos^2{2\theta} - \sin{2\theta} = 2\cos^2{2\theta} - 1 \\$

Therefore our above expression can be simplified as follows:

$\left(\frac{\cos{2\theta} - \cos{4\theta}}{\cos{2\theta} + \cos{4\theta}}\right)^2 =\left(\frac{\cos{2\theta} - \left(2\cos^2{2\theta} - 1\right)} {\cos{2\theta} + \left(2\cos^2{2\theta} - 1\right)}\right)^2= \left(\frac{ \cos{2\theta} - 2\cos^2{2\theta}+ 1 } {\cos{2\theta} + 2\cos^2{2\theta} -1}\right)^2$ $\cos^2{2\theta} = \left(\frac{1}{4}\right)^2 \left(1+\sqrt{5}\right)^2 = \frac{6+2\sqrt{5}}{16} = \frac{3 + \sqrt{5}}{8}$

$\left(\frac{\frac{1}{4}\left(1+\sqrt{5}\right) - \frac{2}{8} \left(3 +\sqrt{5}\right) +1} {\frac{1}{4}\left(1+\sqrt{5}\right) + \frac{2}{8} \left(3 +\sqrt{5}\right) - 1}\right)^2 =\left( \frac{-\frac{2}{4} + 1} {\frac{4+2\sqrt{5}-4}{4}} \right)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}$

Thus $\ m = 1, \ n = 5 \$ and $m+n = 6$

Given : $\cos 2 \theta = \dfrac {1+\sqrt 5}4 = \dfrac {\phi}2$ with $\phi = \dfrac{1+\sqrt 5}2 \quad$ being the "golden ratio"

then $\theta = \dfrac{\pi}{10}$ is a solution which leads to $\quad \tan^2(3\theta) \tan^2(\theta) =.2 \quad$

and $.2 = \dfrac {1}5$ with $1+5= \boxed {6} \quad$ which was the answer sought.

Another Fun Fact ! : $\cos( \dfrac {\pi}5 ) = \dfrac {\phi}2 \quad$

$Cos^2X=\dfrac{Cos2X + 1} 2 =\dfrac{ \frac {1+\sqrt5} 4 + 1} 2=\color{#3D99F6}{\dfrac{5+\sqrt5} 8}\\ Sin^2X=1 - Cos^2X=\color{#3D99F6}{\dfrac{3 - \sqrt5} 8}\\ \color{#3D99F6}{Sin3X=3SinX - 4Sin^3X, ~~~~~~~Cos3X= - 3CosX + 4Cos^3X}\\ \therefore ~Tan3X*TanX=\large \dfrac{3SinX - 4Sin^3X}{ - 3CosX + 4Cos^3X}*\dfrac{ SinX}{CosX}\\ =\dfrac{ Sin^2X}{Cos^2X}* \dfrac{3 - 4Sin^2X}{ - 3 + 4Cos^2X} \\ = \dfrac{\dfrac{3 - \sqrt5} 8 } {\dfrac{5+\sqrt5} 8}* \dfrac{3 - 4\dfrac{3 - \sqrt5} 8}{ - 3 + 4\dfrac{5 + \sqrt5} 8} \\ = \dfrac{ (3 - \sqrt5 ) } {( 5+\sqrt5 )}* \dfrac{6 - (3 - \sqrt5)}{ - 6 + 5 + \sqrt5 } \\ = \dfrac{9 - 4}{\sqrt5*( \sqrt5+1 )*(-1 + \sqrt5)}=\dfrac 4 {4*\sqrt5}.\\ \therefore~Tan^2 3X*Tan^2 X=\dfrac 1 5. ~~\implies m+n= \Large \color{#D61F06}{6}$

hahahaha, I used my Calculator nad found that theta = 18

expression=tan^2(t).tan^2(3t). =1÷(cot^2 (t).cot^2(3t))=m\n. m=1,n=cot^2 (t)+cot^2 (3t). cos(2t)=1+√5\4. So 2cos^2(t)=5+√5\4. cos ^2(t)=5+√5\8 so sin^2(t)=3-√5\8. cos3t=cos2t.cos t-sin2t.sint. =cos t(cos2t-2sin^2(t)). cos ^2(3t)=cos^2(t).((2cos^2(t)-1)-2+2cos^2(t)) ^2. cos^2(t).(4cos^2(t)-3)^2. =(5+√5)\8×(4×(5+√5)\8-3)^2=5-√5\8. so sin^2 (3t)=3-√5\8. n=((5+√5)÷8) ×(8÷(3-√5)×((5-√5)÷8)×(8÷(3-√5)) n=(5-√5).(5+√5)(3-√5).(3+√5)=20÷4=5. m+n=1+5=6#####