First slide
Trigonometric Identities
Question

If 4nα=π then the value of  tan⁡α tan⁡2α tan⁡3α…tan⁡(2n−1)α is 

Moderate
Solution

tan⁡(2n−1)α=tan⁡(π/2−α)=cot⁡α 

tan⁡(2n−2)α=cot⁡2α  and so on. 

So the required  value is 1. 

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