The graphs of \(y = \tan^{-1} x\) and \(y = \cot^{-1} x\) (principal value branches) meet at the point
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\left(1, \displaystyle\frac{\pi}{4}\right)\)
At a meeting point the two values are equal: \(\tan^{-1} x = \cot^{-1} x = \theta\). The sign of x decides this, by the two ranges \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) and \((0, \pi)\).
For \(x \lt 0\): \(\tan^{-1} x\) is negative while \(\cot^{-1} x\) is greater than \(\frac{\pi}{2}\), so they cannot be equal.
For \(x = 0\): \(\tan^{-1} 0 = 0\) and \(\cot^{-1} 0 = \frac{\pi}{2}\), not equal.
For \(x \gt 0\): both values lie in \(\left(0, \frac{\pi}{2}\right)\), and
\(\tan\theta = x \;\text{ and }\; \cot\theta = x\)
\(\tan\theta = \cot\theta \;\Rightarrow\; \tan^{2}\theta = 1\)
With \(\theta\) in \(\left(0, \frac{\pi}{2}\right)\), \(\tan\theta = 1\), so
\( \displaystyle \theta = \frac{\pi}{4}, \quad x = \tan\frac{\pi}{4} = 1 \)
Hence the graphs meet only at \(\left(1, \frac{\pi}{4}\right)\).
Why the other options are incorrect
Option (1) \(\left(-1, -\frac{\pi}{4}\right)\) — Incorrect. This point is on the \(\tan^{-1}\) graph only; \(\cot^{-1}(-1) = \frac{3\pi}{4}\).
Option (3) \(\left(0, \frac{\pi}{2}\right)\) — Incorrect. This point is on the \(\cot^{-1}\) graph only; \(\tan^{-1} 0 = 0\).
Option (4) The graphs do not meet — Incorrect. \(\tan^{-1} 1 = \cot^{-1} 1 = \frac{\pi}{4}\), so they meet at \(x = 1\).