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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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## QuestionMathsClass 11

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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## 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
$$=\tan (\dfrac{\pi }{4n})\tan (\dfrac{\pi }{2n})\tan (\dfrac{3\pi }{4n})\ldotp \ldotp \ldotp \tan (n\times \dfrac{\pi }{4n})\ldotp \ldotp \ldotp \cot (\dfrac{\pi }{2n})\cot (\dfrac{\pi }{4n})$$
$$=\tan \dfrac{\pi }{4}$$
$$=1$$