If xsin(a+y)+sinacos(a+y)=0,provethatdydx=sin2(a+y)sina
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Answer:
The Proof is given in the Explanation.
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.
The Proof is given in the Explanation.
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.
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