WBJEE · Maths · Inverse Trigonometric Functions
The possible values of \(x\), which satisfy the trigonometric equation
\(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\) are
- A \(\pm \frac{1}{\sqrt{2}}\)
- B \(\pm \sqrt{2}\)
- C \(\pm \frac{1}{2}\)
- D \(\pm 2\)
Answer & Solution
Correct Answer
(A) \(\pm \frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
We have, \(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\) \(\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]=\frac{\pi}{4}\)…
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