A) (a) \[\frac{\sin a}{\sin a.{{\sin }^{2}}(a+y)}\]
B) (b) \[\frac{{{\sin }^{2}}(a-y)}{\sin a}\]
C) (c) \[\sin a.{{\sin }^{2}}(a+y)\]
D) (d) \[\frac{{{\sin }^{2}}(a+y)}{\sin a}\]
Correct Answer: D
Solution :
[d] \[\because siny=x.sin\left( a+y \right)\] \[\Rightarrow x=\frac{siny}{sin\left( a+y \right)}\] Differentiating w.r.t. y, we have \[\frac{dx}{dy}=\frac{sin\left( a+y \right).cosy-siny.cos\left( a+y \right)}{\{sin{{(a+y)}^{2}}\}}=\frac{sin\left( a+y-y \right)}{{{\sin }^{2}}(a+y)}\] \[\Rightarrow \frac{dx}{dy}=\frac{si{{n}^{2}}\left( a+y \right)}{\sin a}\] Hence, option [d] is correct.
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