JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If the sum of all the solutions of \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}\) \(-1 < x < 1,x \neq 0,\) is \(\alpha-\frac{4}{\sqrt{3}}\), then \(\alpha\) is equal to \(..........\).
- A \(4\)
- B \(2\)
- C \(6\)
- D \(8\)
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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