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If Y = Tan − 1 ( Sin X + Cos X Cos X − Sin X ) , Then D Y D X is Equal to (A) 1 2 - Mathematics

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प्रश्न

If \[y = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right), \text { then  } \frac{dy}{dx}\] is equal to ___________ .

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उत्तर

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\[\text { We have, y } = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{1 + \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right)^2}\frac{d}{dx}\left( \frac{\sin x + \cos x}{\cos x - \sin x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}\left[ \frac{\left( \cos x - \sin x \right)\frac{d}{dx}\left( \sin x + \cos x \right) - \left( \sin x + \cos x \right)\frac{d}{dx}\left( \cos x - \sin x \right)}{\left( \cos x - \sin x \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}\left[ \frac{\left( \cos x - \sin x \right)\left( \cos x - \sin x \right) - \left( \sin x + \cos x \right)\left( - \sin x - \cos x \right)}{\left( \cos x - \sin x \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}\left[ \frac{\left( \cos x - \sin x \right)\left( \cos x - \sin x \right) + \left( \sin x + \cos x \right)\left( \sin x + \cos x \right)}{\left( \cos x - \sin x \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2} \times \frac{\left( \cos x - \sin x \right)^2 + \left( \sin x + \cos x \right)^2}{\left( \cos x - \sin x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = 1\]

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अध्याय 11: Differentiation - Exercise 11.10 [पृष्ठ १२२]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.10 | Q 33 | पृष्ठ १२२

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