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Question
If \[\sin y = x \cos \left( a + y \right), \text { then } \frac{dy}{dx}\] is equal to ______________ .
Options
\[\frac{\cos^2 \left( a + y \right)}{\cos a}\]
\[\frac{\cos a}{\cos^2 \left( a + y \right)}\]
\[\frac{\sin^2 y}{\cos a}\]
none of these
Solution
\[\frac{\cos^2 \left( a + y \right)}{\cos a}\]
We have,
\[ \Rightarrow \cos y\frac{dy}{dx} = 1 \times \cos\left( a + y \right) - x \sin\left( a + y \right)\frac{d}{dx}\left( a + y \right)\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \cos\left( a + y \right) - x \sin\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow \cos y\frac{dy}{dx} + x \sin\left( a + y \right)\frac{dy}{dx} = \cos\left( a + y \right)\]
\[ \Rightarrow \left[ \cos y + x \sin\left( a + y \right) \right]\frac{dy}{dx} = \cos\left( a + y \right)\]
\[ \Rightarrow \left[ \cos y + \frac{\sin y}{\cos\left( a + y \right)} \times \sin\left( a + y \right) \right]\frac{dy}{dx} = \cos\left( a + y \right) .............\binom{ \because \sin y = x \cos\left( a + y \right)}{ \because x = \frac{\sin y}{\cos\left( a + y \right)}}\]
\[ \Rightarrow \left[ \frac{\cos\left( a + y \right) \cos y + \sin y \sin\left( a + y \right)}{\cos\left( a + y \right)} \right]\frac{dy}{dx} = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{\cos\left( a + y - y \right)}{\cos\left( a + y \right)} \times \frac{dy}{dx} = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\cos a}\]
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