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If Cos Y = X Cos ( a + Y ) , Where Cos a ≠ ± 1 , Prove that D Y D X = Cos 2 ( a + Y ) Sin a ? - Mathematics

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प्रश्न

\[\text{ If }\cos y = x\cos\left( a + y \right),\text{  where } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?
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उत्तर

\[\cos y = x\cos\left( a + y \right)\]
\[ \Rightarrow - \sin y\frac{dy}{dx} = \cos\left( a + y \right) - x\sin\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow - \sin y\frac{dy}{dx} + x\sin\left( a + y \right)\frac{dy}{dx} = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\cos y}{\cos\left( a + y \right)}\sin\left( a + y \right) - \sin y \right] = \cos\left( a + y \right) \left[ \because x = \frac{\cos y}{\cos\left( a + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\cos y\sin\left( a + y \right) - \sin y\cos\left( a + y \right)}{\cos\left( a + y \right)} \right] = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\sin\left( a + y - y \right)}{\cos\left( a + y \right)} \right] = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\sin a}{\cos\left( a + y \right)} \right] = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\]

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पाठ 11: Differentiation - Exercise 11.05 [पृष्ठ ९०]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.05 | Q 57 | पृष्ठ ९०

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