# `tan(cos^-1((x + 1)/2)) = sqrt(4 - (x + 1)^2)/(x + 1)` Verfiy the identity.

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You need to rremember that `tan^2 theta+ 1 = 1/(cos^2 theta)` , hence, replacing `cos^(-1) ((x+1)/2)` by `alpha` , yields:

`cos^(-1) ((x+1)/2) = alpha => cos(cos^(-1) ((x+1)/2)) = cos alpha`

`(x+1)/2 = cos alpha`

Raising to square both sides, yields:

`(x+1)^2/4 = cos^2 alpha =>1/(cos^2 alpha) = 4/((x+1)^2)`

But `1/(cos^2 alpha) = tan^2 alpha + 1` , hence `4/((x+1)^2) = tan^2 alpha + 1.`

`tan^2 alpha = 4/((x+1)^2) - 1`

`tan^2 alpha = (4 - (x+1)^2)/((x+1)^2)`

You need to take square root both sides, such that:

`tan^2 alpha = (4 - (x+1)^2)/((x+1)^2)`

`tan alpha = sqrt((4 - (x+1)^2)/((x+1)^2))`

`tan alpha = sqrt((4 - (x+1)^2)/((x+1)^2))<br>`

**Hence, verifying the given identity `tan(arccos((x+1)/2)) = (sqrt(4-(x+1)^2))/(x+1)` yields that it is true.**

Taking (x+1) as adjacent side and 2 as hypotenuse,we get the opposite side as `sqrt(2^2-(x+1)^2)` .

`tan(cos^-1((x+1)/2))` =` tan(tan^-1(sqrt(4-(x+1)^2)/(x+1)))` = `sqrt(4-(x+1)^2)/(x+1).`

Let `B = cos^-1[(x+1)/2] `

`Cos B = (x+1)/2 `

`(cos B)^2 = (x+1)^2/4 `

`( sin B)^2 = 1- (x+1)^2/4= [ 4-(x+1)^2]/4 `

`sin B = sqrt( [4-(x+1)^2]/4) `

`= sqrt[ 4-(x+1)^2] /2 `

`Tan B= sinB /cosB= (sqrt[ 4-(x+1)^2] /2) / [(x+1)/2] `

`= sqrt[ 4-(x+1)^2] / (x+1) `