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