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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

\(\frac{x}{2}\) के सापेक्ष \(\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)\), जहाँ \(\left(x \in\left(0, \frac{\pi}{2}\right)\right)\) का अवकलज है

  1. A \(2\)
  2. B \(\frac{1}{2}\)
  3. C \(1\)
  4. D \(\frac{2}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

Given \(y = {\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)\) \( \Rightarrow y = {\tan ^{ - 1}}\left( {\frac{{\tan x - 1}}{{\tan x + 1}}} \right)\) \( \Rightarrow y = - {\tan ^{ - 1}}\left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right)\)…
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