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this is how i did it
(i)let PN= y
the triangle PTN is equal to tanx so
0.5x(y x TN)= tanx
TN=2tanx/y
dy/dx = y/ (2tanx/y)
=y x y/2(sinx/cosx)
=y^2 x 1/2(cosx/sinx)
=1/2 y^2 cotx
(ii) dy/dx =y^2/2 x cosx/sinx
2/y^2 dy= cosx/sinx dx (integrate this)
-2y^-1 =ln(sinx) +c
apply y=2 x=1/6 pi
c=-1-ln(1/2)
put it in the equation and get y=2/(1-ln2sinx)
hope you get it
Thank you so much. I was missing just a spot and the whole question seemed to go wrong.