Q109 / 180 · KCET 2018 · Mathematics
If \( \cos y=x \cos (a+y) \) with \( \cos a \neq \pm 1 \), then \( \frac{d y}{d x} \) is equal to
- A\( \frac{\sin a}{\cos ^{2}(a+y)} \)
- Correct answerB\( \frac{\cos ^{2}(a+y)}{\sin a} \)
- C\( \frac{\cos a}{\sin ^{2}(a+y)} \)
- D\( \frac{\cos ^{2}(a+y)}{\cos a} \)
Solution
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
\[ \begin{array}{l} 1=\frac{(-\cos (a+y) \sin y+\sin (a+y) \cos y)}{\cos ^{2}(a+y)} \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin (a+y-y)}=\frac{\cos ^{2}(a+y)}{\sin a} \end{array} \]
Hence, the correct answer is option: 'B'.