Find value of $\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$ My attempt :
$$\dfrac{\left(1+\tan^2\dfrac{5\pi}{12}\right)\left(1-\tan^2\dfrac{\pi}{12}\right)}{\tan\dfrac{\pi}{12}\tan\dfrac{5\pi}{12}}$$
Change into variable form
$$\dfrac{(1+a^2)(1-b^2)}{ab}$$
$$\dfrac{1+a^2-b^2-a^2b^2}{ab}$$
I'm stuck here also I don't think this is the correct way.
 A: $$\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$$
$$=\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{5\pi}{12}}$$
$$=\dfrac{4}{\left(\dfrac{2\tan\frac{5\pi}{12}}{(1+\tan^2\frac{5\pi}{12})}\right)\left(\dfrac{2\tan\frac{\pi}{12}}{{1-\tan^2\frac{\pi}{12}}}\right)}$$
Use trig. identity, $\frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta$, $\frac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$,
$$=\dfrac{4}{\left(\sin\left(2\frac{5\pi}{12}\right)\right)\left(\tan\left(2\frac{\pi}{12}\right)\right)}$$
$$=\dfrac{4}{\left(\sin\frac{5\pi}{6}\right)\left(\tan\frac{\pi}{6}\right)}$$
$$=\dfrac{4}{\left(\frac12\right)\left(\frac{1}{\sqrt 3}\right)}$$
$$=8\sqrt3$$
A: We can write
$$ F=\frac{(1+\tan^2(5\pi/12))(1-\tan^2(\pi/12))}{\tan(5\pi/12) \tan(\pi/12)}$$
$$\implies F=\frac{2}{\frac{2\tan(\pi/12)}{1-\tan^2(\pi/12)}}\frac{\sec^2(5\pi/12)}{\tan(5\pi/12)}.$$
$$F=2 \cot (\pi/6) \frac{\csc^2(\pi/12)}{\cot(\pi/12)}=4 \sqrt{3} \csc(\pi/6)=8\sqrt{3}$$
