what is tan(tan^-1 2+tan^-1 1)?

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You need to find `tan(tan^-1 2 + tan^-1 1),` hence you need to use the formula:

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

`tan(tan^-1 2 + tan^-1 1) =(tan (tan^-1 1) + tan (tan^-1 2))/(1 - tan (tan^-1 1)*tan (tan^-1 2))`

You need to remember that `tan(tan^-1 a) = a` such that:

`tan(tan^-1 2 + tan^-1 1) =(1 + 2)/(1 - 1*2)`

`tan(tan^-1 2 + tan^-1 1) = 3/(1 - 2) =gt tan(tan^-1 2 + tan^-1 1) = -3.`

**Hence, evaluating the value of tangent yields: `tan(tan^-1 2 + tan^-1 1) = -3.` **

tan(tan⁻¹2+tan⁻¹1)

**⇒ tan (A+B) = (tanA + tanB) / (1 - tanA.tanB)**

= [tan(tan⁻¹2) + tan(tan⁻¹1)] / [1-tan(tan⁻¹2).tan(tan⁻¹1)]

**⇒ tan(tan⁻¹x) = x**

=(2+1)/(1-2*1)

= 3/-1

=** -3**

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