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

### 1 Answer | Add Yours

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`

..................................................

`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.` **