If `x=e^(cos2t)`and `y=e^(sin2t)`, prove that `(dy)/(dx)=-(ylogx)/(xlogy)`

If `x=e^(cos2t)`and `y=e^(sin2t)`, prove that `(dy)/(dx)=-(ylogx)/(xlo

1.	If `x=e^(cos2t)`and `y=e^(sin2t)`, prove that `(dy)/(dx)=-(ylogx)/(xlogy)`
Answer» `x=e^(cos2t)and y=e^(sin 2t)` ` cos 2t= log x and sin 2t = log y` `therefore" "cos^(2) 2t +sin^(2) 2t = (log x)^(2) + (log y)^(2)` `rArr" "(log x)^(2)+(log y)^(2)=1` Differentiating both sides w.r.t. x, we get `2log x(1)/(x)+2 log y (1)/(y)(dy)/(dx)=0` `rArr" "(dy)/(dx)=(-y log x)/(x log y)`

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If `f(theta) = sin(tan^(-1)(sintheta/sqrt(cos2theta)))`, where `-pi/4 lt theta lt pi/4,` then the value of `d/(d(tantheta)) f(theta)` is