If x = acos t + log tan(t/2)], y=asint find dy/dx

1.	If x = acos t + log tan(t/2)], y=asint find dy/dx
Answer» X = a(cos∅ +Iog(tan∅/2))y = a sin∅ dx/d∅ = a( -sin∅ +1/tan∅/2 sec²∅/2.1/2) = a{ -sin∅ + (1+tan²∅/2)/2tan∅/2}=a( -sin∅ + cosec∅) =a( 1-sin²∅)/sin∅ =acot∅.cos∅ --------(1) dy/d∅ = acos∅ ------(2) divide (2) /(1) dy/dx = 1/cot∅ dy/dx = tan∅ differentiate wrt x d²y/dx² = sec²∅.d∅/dx d²y/dx² =sec²∅.{ 1/dx/d∅} d²y/dx² = sec²∅.atan∅.sec∅= a.sec³∅.tan∅

Answer»

X = a(cos∅ +Iog(tan∅/2))y = a sin∅

dx/d∅ = a( -sin∅ +1/tan∅/2 sec²∅/2.1/2)

= a{ -sin∅ + (1+tan²∅/2)/2tan∅/2}=a( -sin∅ + cosec∅) =a( 1-sin²∅)/sin∅ =acot∅.cos∅ --------(1)

dy/d∅ = acos∅ ------(2)

divide (2) /(1)

dy/dx = 1/cot∅

dy/dx = tan∅

differentiate wrt x d²y/dx² = sec²∅.d∅/dx

d²y/dx² =sec²∅.{ 1/dx/d∅}

d²y/dx² = sec²∅.atan∅.sec∅= a.sec³∅.tan∅

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