For any integer $m>1$ evaluate the value of $P$:
$$P=\prod_{k=1}^{m-1} \cot{\dfrac{k\pi}{2m}}=\dfrac{\prod_{k=1}^{m-1} \cos{\dfrac{k\pi}{2m}}}{\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{2m}}}=?$$
I don't know how to proceed using complex number's basic properties (like $n$ roots of unity on polygon inscribed in unit circle) without contour integration or advance algebra… I'm class 12th student… I attempted the problem as follows:
My attempt:
we know that
$$\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{m}}=\dfrac{m}{2^{m-1}}$$
$$2^{m-1}\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{2m}} × \cos\dfrac{k\pi}{2m}=\dfrac{m}{2^{m-1}}$$
$$\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{2m}} × \cos\dfrac{k\pi}{2m}=\dfrac{m}{2^{2(m-1)}} $$ but after this step I got stuck because we want ratio of product of cosines and product of sines… and I got product of sines and cosines. I don't know how to relate further… I also tried telescoping series but it didn't work either… Any help will be appreciated, thank you…
Any other method is also welcome
by hint of Gerry myerson i am able to solve this question
we know $cot\left(\dfrac{\pi}{2}-\theta\right)=tan\theta\implies cot\dfrac{(m-1)\pi}{2m}=tan\dfrac{\pi}{2m};cot\dfrac{(m-2)\pi}{2m}=tan\dfrac{2\pi}{2m};....... $
$\implies P=\prod_{k=1}^{m-1} cot{\dfrac{k\pi}{2m}}=cot\dfrac{\pi}{2m}.cot\dfrac{2\pi}{2m}.cot\dfrac{3\pi}{2m}........cot\dfrac{(m-1)\pi}{2m}$
coupling last term with first term ;
$P=\left(cot\dfrac{\pi}{2m}tan\dfrac{\pi}{2m}\right).\left(cot\dfrac{2\pi}{2m}.tan\dfrac{2\pi}{2m}.\right)....\left(cot\dfrac{\dfrac{(m-1)}{2}\pi}{2m}.tan\dfrac{{\dfrac{(m-1)}{2}}\pi}{2m}\right)=1$
$\implies P=1$