# Cos(tan^-1(x))

Cos (tan-1 x)

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

If you need to prove the identity, then here are the steps.

First we will assume that arctanx = A

==> tanA = x

But we know that tanA = `sinA/cosA ` = opposite side/adjacent side= `x/1` Then we can assume that we have aright angle...

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

If you need to prove the identity, then here are the steps.

First we will assume that arctanx = A

==> tanA = x

But we know that tanA = `sinA/cosA ` = opposite side/adjacent side= `x/1` Then we can assume that we have aright angle triangle such that the sides are x and 1.

Then, the hypotenuse is `h= sqrt(1^2+x^2)= sqrt(1+x^2).`

Now we need to find `cos(tan^-1x)`

`cos(tan^-1 x) = cos(A) =(adjacent)/h = 1/sqrt(1+x^2)` ==> `cos(tan^-1 x) = 1/sqrt(1+x^2)`

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