If siny = xsin(a + y)siny=xsin(a+y). Then how will you prove that dy/dx = (sin^2(a + y))/sinadydx=sin2(a+y)sina??

1 Answer
Jul 24, 2017

The Proof is given in the Explanation.

Explanation:

siny=xsin(a+y).siny=xsin(a+y).

:. x=siny/sin(a+y).

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

dx/dy=[sin(a+y)*d/dy{siny}-siny*d/dx{sin(a+y)}]/sin^2(a+y),

={sin(a+y)cosy-sinycos(a+y)*d/dy(a+y)}/sin^2(a+y),..."[The Chain Rule],"

={sin(a+y)cosy-sinycos(a+y)}/sin^2(a+y),

=sin{(a+y)-y}/sin^2(a+y),

=sina/sin^2(a+y).

rArr dy/dx=1/{dx/dy}=sin^2(a+y)/sina.

Enjoy Maths.!