Question

Find `"dy"/"dx"` if : x = cosec²θ, y = cot³θ at θ= `pi/(6)`

Answer 1

x = cosec²θ, y = cot³θ
Differentiating x and y w.r.t. θ, we get
`"dx"/"dθ" = "d"/"dθ"("cosec"θ)^2 = 2"cosec"θ."d"/"dθ"("cosec"θ)`
= 2cosecθ(– cosecθ cotθ)
= – 2cosec²θ cotθ
and
`"dy"/"dθ" = "d"/"dθ"(cotθ)^3 = 3cot^2θ."d"/"dθ"(cotθ)`
= 3cot²θ.(–cosec²θ)
= –3cot²θ.cosec²θ

∴ `"dy"/"dx" = (("dy"/"dθ"))/(("dx"/"dθ")) = (-3cot^2θ."cosc"^2θ)/(-2"cosec"^2θ.cotθ)`
= `(3)/(2)cotθ`

∴ `(dy/dx)_("at"  θ = pi/6)`

= `(3)/(2)cot  pi/(6)`

= `(3sqrt(3))/(2)`.