If y=log [square root of (1-cos x /1+cos x)], find dy/dx.
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y= log [sq.root of (1-cosx / 1+cosx)]
1-cosx x 1-cosx = (1-cosx)^2
1+cosx 1+cosx 1 - cos^2x
= (1-cosx)^2
sin^2x
on taking square root we get
= 1- cosx = cosecx - cot x
sinx
now, y = log(cosecx - cotx)
dy/dx = 1 / (cosecx - cotx) x (-cosec^2x - logsinx)
= (-cosec^2x - logsinx) / (cosecx - cotx)
note: differentiate part alone doubt whether it is right or wrong
1-cosx x 1-cosx = (1-cosx)^2
1+cosx 1+cosx 1 - cos^2x
= (1-cosx)^2
sin^2x
on taking square root we get
= 1- cosx = cosecx - cot x
sinx
now, y = log(cosecx - cotx)
dy/dx = 1 / (cosecx - cotx) x (-cosec^2x - logsinx)
= (-cosec^2x - logsinx) / (cosecx - cotx)
note: differentiate part alone doubt whether it is right or wrong
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