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If \( \theta =\frac{\pi }{4n}\) then the value of \( tan \theta\;tan \left(2\theta \right) tan \left(3\theta \right).... tan \left(\right(2n-1\left)\theta \right)\) is
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If \( \theta =\frac{\pi }{4n}\) then the value of \( tan \theta\;tan \left(2\theta \right) tan \left(3\theta \right).... tan \left(\right(2n-1\left)\theta \right)\) is

Answer

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Solution

To find: \( tan \theta\;tan \left(2\theta \right) tan \left(3\theta \right).... tan \left(\right(2n-1\left)\theta \right)\) when \( \theta =\frac{\pi }{4n}\) 
\(=\tan (\dfrac{\pi }{4n})\tan (\dfrac{2\pi }{4n})\tan (\dfrac{3\pi }{4n})\ldotp \ldotp \ldotp \tan ((2n−2)\times \dfrac{\pi }{4n})\tan ((2n−1)\times \dfrac{\pi }{4n})\)
\(=\tan (\dfrac{\pi }{4n})\tan (\dfrac{\pi }{2n})\tan (\dfrac{3\pi }{4n})\ldotp \ldotp \ldotp \tan (\dfrac{\pi }{2}−\dfrac{\pi }{2n})\tan (\dfrac{\pi }{2}−\dfrac{\pi }{4n})\)
We know that \(\tan (\dfrac{\pi }{2}−\theta )=\cot \theta\)and \(\tan \theta \times \cot \theta =1\)as \(\tan \theta =\dfrac{1}{\cot \theta }\)
\(=\tan (\dfrac{\pi }{4n})\tan (\dfrac{\pi }{2n})\tan (\dfrac{3\pi }{4n})\ldotp \ldotp \ldotp \cot (\dfrac{\pi }{2n})\cot (\dfrac{\pi }{4n})\)
Only middle term is left,
Total terms \(2n-1\)so middle term will be \(\dfrac{2n−1+1}{2}={n}^{th}\)term
\(=\tan (\dfrac{\pi }{4n})\tan (\dfrac{\pi }{2n})\tan (\dfrac{3\pi }{4n})\ldotp \ldotp \ldotp \tan (n\times \dfrac{\pi }{4n})\ldotp \ldotp \ldotp \cot (\dfrac{\pi }{2n})\cot (\dfrac{\pi }{4n})\)
\(=\tan \dfrac{\pi }{4}\)
\(=1\)
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