यदि `x=(sin^(3)t)/(sqrt(cos2t)),y=(cos^(3)t)/(sqrt(cos2t))`, तब `(dy)/(dx)` ज्ञात कीजिए।

Explanation

यहाँ, `x=(sin^(3)t)/(sqrt(cos2t))`
`sqrt(cos2t).(3sin^(2)tcost)`
`therefore(dx)/(dt)=(-sin^(3)t(1/(2sqrt(cos2t)).(-2sin2t)))/((sqrt(cos2t))^(2))`
`=(cos2t(3sin^(2)tcost)+sin^(3)tsin2t)/((sqrt(cos2t)))`
`=(sin^(2)t(3costcos2t+sintsin2t))/((cos2t)^(3//2))`
`=(sin^(2)t[2costcos2t+(cos2tcost+sin2tsint)])/((cos2t)^(3//2))`
`=(sin^(2)t[2costcos2t+cos(2t-t)])/((cos2t)^(3//2))`
`=(sin^(2)t[2costcos2t+cost])/((cos2t)^(3//2))`
`rArr(dx)/(dt)=(sin^(2)tcost(2cos2t+1))/((cos2t)^(3//2))`
और `y=(cos^(3)t)/(sqrt(cos2t))`
`therefore(dy)/(dt)=(sqrt(cos2t)(-3cos^(2)tsint)-cos^(3)t(1/(2sqrt(cos2t)).(-2sin2t)))/((sqrt(cos2t))^(2))`
`=(cos2t(-3cos^(2)tsint)+cos^(3)t.sin2t)/((sqrt(cos2t))^(3))`
`=(cos^(2)t[-3cos2tsint+costsin2t])/((cos2t)^(3//2))`
`=(cos^(2)t[-2cos2tsint+(sin2tcost-cos2tsint)])/((cos2t)^(3//2))`
`=(cos^(2)t[-2cos2tsint+sin(2t-t)])/((cos2t)^(3//2))`
`rArr(dy)/(dt)=(cos^(2)tsint(1-2cos2t))/((cos2t)^(3//2))`
अब, `(dy)/(dx)=(dy//dt)/(dx//dt)`
`=(cos^(2)tsint(1-2cos2t))/((cos2t)^(3//2))xx((cos2t)^(3//2))/(sin^(2)tcost(1+2cos2t))`
`=(cos(1-2cos2t))/(sint(1+2cos2t))`
`=cot[(1-2((1-tan^(2)t)/(1+tan^(2)t)))/(1+2((1-tan^(2)t)/(1+tan^(2)t))]`
`=1/(tant).(-1+3tan^(2)t)/((3-tan^(2)t))`
`=-((1-3tan^(2)t))/((3tant-tan^(3)t))`
`=-1/(tan3t)`
`=-cot3t`