Question

x = 3cosθ – 2cos³θ, y = 3sinθ – 2sin³θ

Answer 1

दिया गया है: x = 3 cosθ – 2 cos³θ और y = 3sinθ – 2 sin³θ.

दोनों प्राचलिक फलनों को अलग करना w.r.t. θ

`"dx"/("d"theta) = -3 sin theta - 6cos^2theta * "d"/("d"theta) (cos theta)`

= – 3 sin θ – 6 cos²θ . (– sin θ)

= – 3 sin θ + 6 cos²θ . sin θ

`"dy"/("d"theta) = 3 os theta - 6 sin^2theta * "d"/("d"theta) (sin theta)`

= = 3 cos θ – 6 sin²θ . cos θ क्योंकि

∴ `"dy"/"dx" = ("dy"/("d"theta))/("dx"/("d"theta))`

= `(3 cos theta - 6 sin^2theta cos theta)/(-3sin theta + 6cos^2 theta * sin theta)`

⇒ `"dy"/"dx" = (cos theta (3 - 6sin^2theta))/(sintheta(-3 + 6 cos^2 theta))`

= `(costheta[3 - 6(1 - cos^2theta)])/(sintheta[-3 + 6cos^2theta])`

= `cot theta ((3 - 6 + 6 cos^2 theta)/(-3 + 6 cos^2theta))`

= `cot theta ((-3 + 6 cos^2theta)/(-3 + 6 cos^2 theta))`

= cot θ

∴ `"dy"/"dx"` = cot θ.