If 4nalpha=π then the value of tan alpha tan2alpha tan3alpha_ _ _ _ tan(2n-1)alpha is equal to

i) There are (2n - 1) factors; Of these let us group a pair of products, selecting the corresponding parts from either end; thus one middle will be left out without any pair for the same. ii) So the arrangement for them is: P = [cot(2na - a)*cot(a)]*[cot(2na - 2a)*cot(2a)]*[cot(2na - 3a)*cot(3a)]*.....*cot(na) iii) When 4na = π, na = π/4; so cot(na) = cot(π/4) = 1 cot(2na - a) = cot(π//2 - a) = tan(a) So cot(2na - a)cot(a) = tan(a)cot(a) = 1 Similarly cot(2na - 2a)cot(2a) = cot(π//2 - 2a)cot(2a) = tan(2a)cot(2a) = 1 Similarly every pair will simplifies to 1 iv) Substituting all these in (ii). above, P = 1 x 1 x 1 x 1 x 1 ....... x 1 = 1 Thus the answer is 1.