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Knowing that tan x = cot ( 9 0 − x ) , we can rewrite the given expression as
tan 1 ⋅ tan 2 ⋅ . . . ⋅ tan 4 5 ⋅ cot ( 9 0 − 4 6 ) ⋅ cot ( 9 0 − 4 7 ) ⋅ . . . ⋅ cot ( 9 0 − 8 9 )
As cot x ⋅ tan x = 1 , then we can pair the terms as follows
tan 1 ⋅ cot 1 ⋅ tan 2 ⋅ cot 2 ⋅ . . . ⋅ tan 4 4 ⋅ cot 4 4 ⋅ tan 4 5
Which is equal to 1 ⋅ 1 ⋅ . . . ⋅ 1 ⋅ tan 4 5 = tan 4 5 = 1