Question

If y=(1+cosx)/sinx then prove that (1-cosx)^2d2y/dx2=sinx

Answer 1

\[y=\frac{ 1+\cos x }{\sin x }\times \frac{ 1-\cos x }{ 1-\cos x }=\frac{ 1-\cos ^{2}x }{ \sin x \left( 1-\cos x \right)}=\frac{ \sin ^{2}x }{\sin x \left( 1-\cos x \right) }\]
\[y=\frac{ \sin x }{1-\cos x },\frac{ dy }{dx }=\frac{ \left( 1-\cos x \right)\cos x-\sin x \left( \sin x \right) }{\left( 1-\cos x \right)^{2} }\]
\[\frac{ dy }{ dx }=\frac{ \cos x-\left( \cos ^{2}x+\sin ^{2}x \right) }{ \left( 1-\cos x \right)^{2} }\]
\[\frac{ dy }{dx }=\frac{ -1\left( 1-\cos x \right) }{ \left( 1-\cos x \right)^{2} }=\frac{ -1 }{ \left( 1-\cos x \right) }=-1\left( 1-\cos x \right)^-1\]

Answer 2

differentiate again w.r.t. x and get the solution.