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Differentiate Tan − 1 ( a + B Tan X B − a Tan X ) ? - Mathematics

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प्रश्न

Differentiate \[\tan^{- 1} \left( \frac{a + b \tan x}{b - a \tan x} \right)\] ?

उत्तर

\[\text{ Let, y } = \tan^{- 1} \left[ \frac{a + b \tan x}{b - a \tan x} \right]\]

\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a + b \tan x}{b}}{\frac{b - a \tan x}{b}} \right]\]

\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a}{b} + \tan x}{1 - \frac{a}{b}\tan x} \right]\]

\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\tan\left( \tan^{- 1} \frac{a}{b} \right) + \tan x}{1 - \tan\left( \tan^{- 1} \frac{a}{b} \right) \times \tan x} \right]\]

\[ \Rightarrow y = \tan^{- 1} \left[ \tan\left( \tan^{- 1} \frac{a}{b} + x \right) \right]\]

\[ \Rightarrow y = \tan^{- 1} \left( \frac{a}{b} \right) + x\]

Differentiate it with respect to x,

\[\frac{d y}{d x} = 0 + 1\]

\[ \therefore \frac{d y}{d x} = 1\]

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अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.03 | Q 27 | पृष्ठ ६३

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