Find dy/dx, If X=a[cos t +log(tan(t/2))] …
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Posted by Aman Badhwar 4 years, 1 month ago
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Gaurav Seth 4 years, 1 month ago
x = a(cost + log tant/2)
differentiate with respect to t,
dx/dt = d{a(cost + logtan t/2)}/dt
= a[d(cost)/dt + d(logtan t/2)/dt]
= a[ -sint + 1/tan t/2 × sec²t/2× 1/2]
= a [ -sint + 1/{2sin t/2.cos t/2} ]
we know, 2sinA.cosA = sin2A , use it here
= a [ -sint + 1/sint]
= a { 1 - sin²t}/sint
= acos²t/sint
= acott. cost .......(1)
again, y = asint
dy/dt = acost .......(2)
now, dy/dx = {dy/dt}/{dx/dt}
= {acost}{acott.cost}
= 1/cott
hence, dy/dx = tan t
1Thank You