`(dy)/(dx)` ज्ञात कीजिए|
`x=(sin^(3)t)/(sqrt(cos 2t)), y=(cos^(3)t)/(sqrt(cos2t))`

Explanation

`x=(sin^(3)t)/(sqrt(cos 2t))`
`rArr" "(dx)/(dt)=(sqrtcos(2t).3 sin^(2)t cos t-sin^(3)t(d)/(dt)sqrt(cos2t))/((sqrt(cos2t))^(2))`
`=(sqrt(cos2t).3sin^(2)t cos t-sin^(3)t.((-2sin2t)/(2sqrt(cos2t))))/(cos2t)`
`=(cos2t.3sin^(2)t cos t+sin^(3)t sin 2t)/(cos2tsqrt(cos2t))`
`=(sin^(2)t[3cos t cos 2t+sin t sin 2t])/(cos2tsqrt(cos2t))`
`=(sin^(2)t[3cost(1-2sin^(2)t)+sint.2sint cost])/(cos2tsqrt(cos2t))`
`=(sin^(2)t cost(3-5sin^(2)t))/(cos2tsqrt(cos2t))`
तथा `y=(cos^(3)t)/(sqrt(cos2t))`
`rArr" "(dy)/(dx)=(sqrt(cos2t)(d)/(dt)cos^(3)t-cos^(3)t(d)/(dt)sqrt(cos2t))/((sqrt(cos2t))^(2))`
`(sqrt(cos2t).3cos^(2)t(-sint)-cos^(3)t.((-2sin2t))/(2sqrt(cos2t)))/(cos2t)`
`=-((3cos^(2)t sin t.cos2t-sin 2tcos^(3)t))/(cos2tsqrt(cos2t))`
`=-(cos^(2)t[3sint(2 cos^(2)t-1)-2sint cost. cost])/(cos2tsqrt(cos2t))`
`=-(cos^(2)t sin t (4 cos^(2)t-3))/(cos2tsqrt(cos2t))`
अब `(dy)/(dx)=(dy//dt)/(dx//dt)=-(cos^(2)t sin t (4 cos^(2)t-3))/(sin^(2)t cost(3-4 sin^(2)t))`
`=-(cos t(4cos^(2)t-3))/(sint(3-4sin^(2)t))`
`=-(cos3t)/(sin3t)=-cot3t`