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Question
Differentiate \[\tan^{- 1} \left( \frac{a + b \tan x}{b - a \tan x} \right)\] ?
Solution
\[\text{ Let, y } = \tan^{- 1} \left[ \frac{a + b \tan x}{b - a \tan x} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a + b \tan x}{b}}{\frac{b - a \tan x}{b}} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a}{b} + \tan x}{1 - \frac{a}{b}\tan x} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\tan\left( \tan^{- 1} \frac{a}{b} \right) + \tan x}{1 - \tan\left( \tan^{- 1} \frac{a}{b} \right) \times \tan x} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \tan\left( \tan^{- 1} \frac{a}{b} + x \right) \right]\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{a}{b} \right) + x\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = 0 + 1\]
\[ \therefore \frac{d y}{d x} = 1\]
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