# How calculate tan(arctan 2-arctan3)?

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### 2 Answers

You should use the following trigonometric identities such that:

`tan(alpha + beta) = (tan alpha + tan beta)/(1 - tan alpha*tan beta)`

`tan(arctan alpha) = alpha`

Substituting `arctan 2` for `alpha` and `arctan 3` for `beta` yields:

`tan(arctan 2+ arctan 3) = (tan(arctan2) + tan(arctan3))/(1 - tan (arctan2)*tan(arctan3))`

`tan(arctan 2+ arctan 3) = (2+3)/(1-2*3)`

`tan(arctan 2+ arctan 3) = 5/(1-6) => tan(arctan 2+ arctan 3) = 5/(-5)`

`tan(arctan 2+ arctan 3) = -1 => {(arctan 2+ arctan 3 = pi - pi/4),(arctan 2+ arctan 3 = 2pi - pi/4):} => {(arctan 2+ arctan 3 = 3pi/4),(arctan 2+ arctan 3 = 7pi/4):}`

**Hence, evaluating the given tangent yields `tan(arctan 2+ arctan 3) = -1.` **

**Sources:**

tan(arctan 2-arctan3):

tan(a+b)=(tana+tanb)/1-tana*tanb....(1)

tanarctan 2=a

tanarctan 3=b

value of a,b put in eq(1)

tan(arctan 2-arctan3):= -1