Question

if $\sf 4n\apha=\pi$

then find the value of :-

$\\ \sf \cot\alpha. \cot 2\alpha. \cot3 \alpha... \cot((2n - 1) \alpha )$
Need full explanation! ​

Answer 1

Given-

$\sf 4n\alpha=\pi$

To Find-

Value of -

$\sf \cot\alpha. \cot 2\alpha. \cot3 \alpha... \cot((2n - 1) \alpha )$

Solution-

$\sf 4n\alpha=\pi$

$\sf n=\frac{π}{4\alpha}$

»Taking general term

$cot(2n-1)\alpha$$\implies$$cot(\frac{\cancel2π}{\cancel4\alpha}-1)\alpha$

$\implies$$cot(\frac{π}{2}-\alpha)$

$\red\implies$$tan\alpha$

and,

$cot(2n-2)\alpha$$\implies$$cot(\frac{\cancel2π}{\cancel4\alpha}-2)\alpha$

$\implies$$cot(\frac{π}{2}-2\alpha)$

$\red\implies$$tan2\alpha$

Also,

$cot(2n-3)\alpha$$\implies$$cot(\frac{\cancel2π}{\cancel4\alpha}-3)\alpha$

$\implies$$cot(\frac{π}{2}-3\alpha)$

$\red\implies$$tan3\alpha$

and it goes on. . .

So, $cot\alpha.cot2\alpha.cot3\alpha. . . .cot(2n−3)\alpha.cot(2n−2)\alpha.cot(2n−1)\alpha$

$\purple\implies$$cot\alpha.cot2\alpha.cot3\alpha. . . . .tan3\alpha.tan2\alpha.tan\alpha$

$\purple\implies$$(cot\alpha.tan\alpha)(cot2\alpha.tan2\alpha)(cot3\alpha.tan3\alpha). . . . .$

$\purple\implies$$1.1.1 . . . . .$

$\red\implies$$\huge1$

Answer 2

⇝ Given :

$\sf 4n\alpha=\pi$

⇝ To Find :

Value of $\\ \sf \cot\alpha. \cot 2\alpha. \cot3 \alpha... \cot\{(2n - 1) \alpha \}$

⇝ Solution :

❒ We Have ,

$4n \alpha = \pi$

$\purple{ \large :\longmapsto  \underline {\boxed{{\pmb{ n = \frac{\pi}{4 \alpha } }} }}}$

❒ Taking some last terms of the expression which we have to find :

$\cot \{(2n - 1) \alpha \}$

★ Putting Value of n :

$= \cot \bigg \{ \bigg(2 \times \dfrac{\pi}{4 \alpha } - 1 \bigg) \alpha \bigg \}$

$= \cot \bigg( \dfrac{ \frac{\pi}{2} - \alpha}{ \alpha } \times \alpha \bigg )$

$= \cot \bigg( \dfrac{ \pi}{2} - \alpha \bigg )$

$= \tan \alpha$

❒ Similarly :

$\cot \{(2n - 2) \alpha \}$

★ Putting Value of n :

$= \cot \bigg \{\bigg( 2 \times \dfrac{\pi}{4 \alpha } \bigg) \alpha \bigg \}$

$= \cot \bigg( \frac{\pi}{2} - \alpha \bigg )$

$= \tan2 \alpha$

Hence,

$\cot\alpha. \cot 2\alpha. \cot3 \alpha... \cot\{(2n - 1) \alpha \}$

$= \cot\alpha. \cot 2\alpha. \cot3 \alpha\: . \: . \: . \: \tan3 \alpha .\tan2 \alpha . \tan \alpha \}$

$= (\cot \alpha \tan .\alpha ). (\cot2 \alpha \tan .2\alpha ). (\cot 3\alpha \tan .3\alpha ) \: . \: . \: . \:$

$= 1.1.1 \: . \: . \: . \:$

$\huge\pink{ \pmb{\underline{\frak{{=1}}}}}$