Question

Sin y= x sin(a+y), prove that dy+dx=sin^2(a+y)+sin a

Answer 1

Explanation:

siny=xsin(a+y).

∴x=sinysin(a+y).

Differentiating w.r.t. y, using the Quotient Rule, we have,

dxdy=sin(a+y)⋅ddy{siny}−siny⋅ddx{sin(a+y)}sin2(a+y),

=sin(a+y)cosy−sinycos(a+y)⋅ddy(a+y)sin2(a+y),...[The Chain Rule],

=sin(a+y)cosy−sinycos(a+y)sin2(a+y),

=sin{(a+y)−y}sin2(a+y),

=sinasin2(a+y).

⇒dydx=1dxdy=sin2(a+y)sina