Question

Find `"dy"/"dx"`, if : `x = cos^-1((2t)/(1 + t^2)), y = sec^-1(sqrt(1 + t^2))`

Answer 1

`x = cos^-1((2t)/(1 + t^2)), y = sec^-1(sqrt(1 + t^2))`
Put t = tanθ.
Then θ =tan^–1t.
∴ x = `cos^-1((2tanθ)/(1 + tan^2θ)), y = sec^-1(sqrt(1 + tan^2 θ))`

∴ x = `cos^-1(sin2θ), y = sec^-1(sqrt(sec^2θ))`

∴ x = `cos^-1[cos(pi/2 - 2θ)], y = sec^-1(secθ)`

∴ x = `pi/2 - 2θ, y = θ`

∴ x =  `pi/(2) - 2tan^-1t, y = tan^-1t`
Differentiating x and y w.r.t. x, we get
`"dx"/"dt" = "d"/"dt"(pi/2) - 2"d"/"dt"(tan^-1t)`

= `0 - 2 xx (1)/(1 + t^2)`

= `(-2)/(1 + t^2)`
and
`"dy"/"dt" = "d"/"dt"(tan^-1t)`

= `(1)/(1 + t^2)`

∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`

= `(((1)/(1 + t^2)))/(((-2)/(1 + t^2))`

=`-(1)/(2)`.