Q101 / 180 · KCET 2015 · Mathematics
Solve for \( x \)
\[
\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x, x>0
\]
\[ \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x, x>0 \]
- A\( \sqrt{3} \)
- B\( 11 \)
- C\( -1 \)
- Correct answerD\( \frac{1}{\sqrt{3}} \)
Solution
Given that, \( \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \)
Taking L.H.S, we have \( \tan ^{-1}\left(\frac{1-x}{1+x}\right) \)
We know that, \( \tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\tan ^{-1} x-\tan ^{-1} y \)
So, \( \tan ^{-1} 1-\tan ^{-1} x=\frac{I}{4}-\tan ^{-1} x \)
So, \( \frac{\Pi}{4}-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x \)
\( \Rightarrow \frac{I I}{4}=\frac{3}{2} \tan ^{-1} x \Rightarrow \tan ^{-1} x=\frac{I I}{6} \)
\( \Rightarrow x=\tan \frac{\Pi}{6} \Rightarrow x=\frac{1}{\sqrt{3}} \)
Taking L.H.S, we have \( \tan ^{-1}\left(\frac{1-x}{1+x}\right) \)
We know that, \( \tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\tan ^{-1} x-\tan ^{-1} y \)
So, \( \tan ^{-1} 1-\tan ^{-1} x=\frac{I}{4}-\tan ^{-1} x \)
So, \( \frac{\Pi}{4}-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x \)
\( \Rightarrow \frac{I I}{4}=\frac{3}{2} \tan ^{-1} x \Rightarrow \tan ^{-1} x=\frac{I I}{6} \)
\( \Rightarrow x=\tan \frac{\Pi}{6} \Rightarrow x=\frac{1}{\sqrt{3}} \)
Hence, the correct answer is option: 'D'.