If n=π4α, then tanα tan2α tan3α.......tan2n−1α is equal to

h=π4αApply tanα.cotα=1Now tanα.tan2π4α−1α=tanα.tanπ2−α=tanα.cotα=1 Similarly remaining all terms.

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If $n = \frac{π}{4 α},$ then $tanα \tan 2 α \tan 3 α . ...... \tan (2 n - 1) α$ is equal to

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Detailed Solution

$h = \frac{π}{4 α}$
Apply $tanα . cotα = 1$
Now $tanα . \tan (2 \frac{π}{4 α} - 1) α$
$\begin{array}{l} = tanα . \tan (\frac{π}{2} - α) \\ = tanα . cotα = 1 \end{array}$
Similarly remaining all terms.

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