KCET · Maths · Limits
Solve for \( x \)
\[
\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x, x>0
\]
- A \( \sqrt{3} \)
- B \(\pi\)
- C \( -1 \)
- D \( \frac{1}{\sqrt{3}} \)
Answer & Solution
Correct Answer
(D) \( \frac{1}{\sqrt{3}} \)
Step-by-step Solution
Detailed explanation
\( \tan^{-1} 1 - \tan^{-1} x = \frac{1}{2} \tan^{-1} x \) \( \frac{\pi}{4} = \frac{3}{2} \tan^{-1} x \)
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