If $4n\alpha =\pi ,$ then the numerical value of $\tan \alpha \tan 2\alpha \tan 3\alpha \mathrm{....}\tan (2n-1)\alpha$ is equal to

$\therefore 4n\alpha =\pi$

$\because \tan (2n-1)\alpha =\tan (2n\alpha -\alpha )=\tan (\frac{\pi }{2}-\alpha )=\cot \alpha$

$\tan \alpha (2n-2)\alpha =\tan (2n\alpha -2\alpha )$

$=\tan (\frac{\pi }{2}-2\alpha )=\cot 2\alpha \mathrm{..........}$

$\therefore \tan \alpha \tan 2\alpha \tan 3\alpha \mathrm{......}\tan (2n-1)\alpha$

$\left(\tan \alpha \tan (2n-1)\alpha \right)\left(\tan 2\alpha \tan (2n-2)\alpha \right)\mathrm{....}\tan n\alpha$

$=\left(\mathrm{11.1....1}\right)\tan n\alpha =\tan n\alpha =\tan \frac{\pi }{4}=1$