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Find dy/dx when x=alogt and y=bsint....

Find `dy/dx` when `x=alogt and y=bsint.`

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The correct Answer is:
1
`x=a[cos t + log tan""(t)/(2)]and y= a sin t`
Differentiating w.r.t. t, we get
`=a[-sin t + (1)/(tan t//2)sec^(2)""(t)/(2)xx(1)/(2)]`
`(dx)/(dt)=a[-sin t+(1)/(2 sin (t//2)cos (t//2))]`
`=a[-sin t +(t)/(sin t)] and (dy)/(dt)=a cos t`
`therefore" "(dy)/(dx)=(dy//dt)/(dx//dt)=(a cos t)/((a cos^(2)t)/(sin t))=tan t`
`"At "t=pi//4, (dy)/(dx)=1`
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