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If U = Cot − 1 √ Tan θ − Tan − 1 √ Tan θ Then , Tan ( π 4 − U 2 ) = (A) √ Tan θ (B) √ Cot θ (C) Tan θ (D) Cot θ - Mathematics

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Question

\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]

Options

  • `sqrt(tantheta`

  • `sqrt(cottheta)`

  •  tan θ

  • cot θ

MCQ
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Solution

(a) `sqrt(tantheta`
Let \[y = \sqrt{\tan\theta}\]
Then, 
\[u = \cot^{- 1} \sqrt{\tan\theta} - \tan^{- 1} \sqrt{\tan\theta}\]
\[ \Rightarrow u = \cot^{- 1} y - \tan^{- 1} y\]
\[ \Rightarrow u = \frac{\pi}{2} - 2 \tan^{- 1} y \left[ \because \tan^{- 1} x + \cot^{- 1} x = \frac{\pi}{2} \right]\]
\[ \Rightarrow 2 \tan^{- 1} y = \frac{\pi}{2} - u \]
\[ \Rightarrow \tan^{- 1} y = \frac{\pi}{4} - \frac{u}{2}\]
\[ \Rightarrow y = \tan\left( \frac{\pi}{4} - \frac{u}{2} \right)\]
\[ \Rightarrow \sqrt{\tan\theta} = \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) \left[ \because y = \sqrt{\tan\theta} \right]\]
\[\]

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Chapter 4: Inverse Trigonometric Functions - Exercise 4.16 [Page 120]

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RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 12 | Page 120

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