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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

Let \(0 < \alpha < 1\), \(\beta = \dfrac{1}{3\alpha}\) and \(\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}\). Then \(6(\alpha + \beta)\) is equal to:

  1. A \(6\)
  2. B \(7\)
  3. C \(8\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(7\)

Step-by-step Solution

Detailed explanation

Given \(\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}\) Applying the formula \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)\), we get: \(\dfrac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)} = \tan\left(\dfrac{\pi}{4}\right) = 1\)…
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