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

જો \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}\)   \(-1 < x < 1,x \neq 0\)ના બધાજ ઉકેલો નો સરવાળો \(\alpha-\frac{4}{\sqrt{3}}\) હોય, તો \(\alpha=...............\).

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

Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

Case \(I: x > 0\) \(\tan ^{-1} \frac{2 x}{1-x^2}+\tan ^{-1} \frac{2 x}{1-x^2}=\frac{\pi}{3}\) \(x=2-\sqrt{3}\) Case \(II:\) \(x < 0\) \(\tan ^{-1} \frac{2 x}{1-x^2}+\tan ^{-1} \frac{2 x}{1-x^2}+\pi=\frac{\pi}{3}\) \(x=\frac{-1}{\sqrt{3}} \Rightarrow \alpha=2\)
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