Question

The derivative of `sin^-1 ((2x)/(1 + x^2))` with respect to `cos^-1 [(1 - x^2)/(1 + x^2)]` is equal to

Options

• 1

• – 1

• 2

• None of these

Answer 1

1

Explanation:

Let s = `sin^-1 ((2x)/(1 + x^2))` and t = `cos^-1 ((1 - x^2)/(1 + x^2))`

We have to find out `(ds)/(dt)`

Putting x = tan θ, we get

s = `sin^-1 [(2 tan theta)/(1 + tan^2 theta)]`

= `sin^-1 (sin 2 theta) = 2theta = 2 tan^-1x`

∴ `(ds)/(dx) = 2/(1 + x^2)`

And t = `cos^-1 ((1 - x^2)/(1 + x^2)) = cos^-1 ((1 - tan^2 theta)/(1 + tan^2 theta))`

= `cos^-1 (cos 2theta) = 2theta = 2 tan^-1x`

∴ `(dt)/(dx) = 2/(1 + x^2)`

∴ `(ds)/(dt) = ((ds)/(dx))/((dt)/(dx)) = 2/(1 + x^2) xx (1+ x^2)/2` = 1