Question

if
$4n \alpha = \pi \\ \cot( \alpha \times) \times \cot(2 \alpha ) \times \cot(3 \alpha ) ....... \cot(2n - 1) =$
is equal to​

Answer 1

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Answer 2

Step-by-step explanation:

Given:

$4n \alpha = \pi \\ :\longrightarrow \bf{\: 2n \alpha = \frac{\pi}{2}}$

Solution:-

★ $\large\bf {\cot \alpha }$

$\tt\cot \alpha . \cot(2n - 1) \alpha = \cot\alpha \cot( \frac{\pi}{2} - \alpha ) \\ \: \: \: \: \: \: \: \: \: \: \: \therefore\bf{ \cot( \frac{\pi}{2} - \alpha ) = \tan \alpha } \\ \\ \: \: \: \: \: \: \: \: : \implies \bf{\cot\alpha . \tan\alpha = 1 \: }$

similarly,

★$\large\bf {\cot 2\alpha }$

$:\implies \cot2\alpha .\cot(2n - 2) \alpha= 1$

★$\large\bf {\cot 3\alpha }$

$:\implies \bf{ \cot3 \alpha . \cot(2n - 3) \alpha = 1}$

★$\large\bf {\cot (2n-1)\alpha}$

$:\implies \bf \cot(n - 1) \alpha . \cot(n + 1) \alpha = 1$

Therefore,

$\sf \cot \alpha \cot2 \alpha \cot3 \alpha ...... \cot(2n - 1) \alpha$

$: \implies \tt {\cot \alpha . \cot(2n - 1)}{\cot2\alpha .\cot(2n - 2) \alpha}$... ..... $\cot(n - 1) \alpha \cot(n + 1) \alpha . \cot \: n \alpha$

$\bf \therefore \cot \: n \alpha = \cot( \frac{\pi}{4} = 1)$

$: \implies \tt 1.1.1....1.1 \\ \\ : \implies \tt 1$