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If \( \cos y=x \cos (a+y) \) with \( \cos a \neq \pm 1 \), then \( \frac{d y}{d x} \) is equal to
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Verified Answer
The correct answer is:
\( \frac{\cos ^{2}(a+y)}{\sin a} \)
Given that \( \cos y=x \cos (a+y) \)
\[
\Rightarrow x=\frac{\cos y}{\cos (a+y)}
\]
Differentiating with respect to \( x \), we get
\[
\begin{array}{l}
1=\frac{(-\cos (a+y) \sin y+\sin (a+y) \cos y)}{\cos ^{2}(a+y)} \frac{d y}{d x} \\
\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin (a+y-y)}=\frac{\cos ^{2}(a+y)}{\sin a}
\end{array}
\]
\[
\Rightarrow x=\frac{\cos y}{\cos (a+y)}
\]
Differentiating with respect to \( x \), we get
\[
\begin{array}{l}
1=\frac{(-\cos (a+y) \sin y+\sin (a+y) \cos y)}{\cos ^{2}(a+y)} \frac{d y}{d x} \\
\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin (a+y-y)}=\frac{\cos ^{2}(a+y)}{\sin a}
\end{array}
\]
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