# Calculate product trigo tan1*tan2*tan*3*tan...*tan88*tan89=

sciencesolve | Certified Educator

You need to remember that `tan alpha = cot(90^o - alpha), ` hence `tan 89^o = cot(90^o - 89^o) = cot 1^o` .

You also need to remember that `cot alpha = 1/tan alpha` , hence, reasoning by analogy yields:

`tan 89^o = cot 1^o = 1/tan 1^o`

`tan 88^o = cot 2^o = 1/tan 2^o`

`tan 87^o = cot 3^o = 1/tan 3^o`

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`tan 46^o = cot 44^o = 1/tan 44^o`

`tan 45^o = cot 45^o = 1`

Hence, substituting `1/tan 1^o`  for tan `89^o` , `1/tan 2^o`  for `tan 88^o` ,...., yields:

`tan 1^o*tan 2^o*...*tan 44^o*tan 45^o*tan 46^o*...*tan 88^o*tan 89^o = tan 1^o*tan 2^o*...*tan 44^o*tan 45^o*(1/tan 44^o)*.....*(1/tan 2^o)*(1/tan 1^o)`

Reducing like terms yields:

`tan 1^o*tan 2^o*...*tan 44^o*tan 45^o*tan 46^o*...*tan 88^o*tan 89^o = tan 45^o = 1`

Hence, evaluating the given product yields `tan 1^o*tan 2^o*...*tan 44^o*tan 45^o*tan 46^o*...*tan 88^o*tan 89^o = 1.`