If `x=sin^3t/(sqrtcos2t), y=cos^3t/sqrt(cos2t)` show that `dy/dx =0 at t=pi/6`

Answer

`-cot 3t`

Explanation

`logx=3logsint-(1)/(2)log cos2t andlogy=3logcost-(1)/(2)log cos 2t`
`rArr(1)/(x).(dx)/(dt)=(3)/(sin t).cost-(1)/(2).(1)/(cos2t)(-2sin 2t),(dy)/(dt).(dy)/(dt)=y.(-3tant +tan2t)`
`rArr(dx)/(dt)=x.[3cot t+tan2t],(dy)/(dx)=y.[-3tant+tan21]`
`rArr(dy)/(dx)=((dy//dt))/((dx//dt))=(y)/(x).((-3tant+tan2t))/((3cot t+tan 2t))=(cot^(3)t.(-3 tan t+(2 tan t)/(1-tan^(2)t)))/(((3)/(tan t)+(2 tant)/(1-tan^(2)t)))`
`=(-(1-3 tan^(2)t))/((3 tant-tan^(3)t))=(-1)/(tan3t)=-cot3t.`