Consider the following statements about the graphs of inverse trigonometric functions (principal value branches):
(A) The graph of \(y = \tan^{-1} x\) passes through the point \(\left(1, \displaystyle\frac{\pi}{4}\right)\).
(B) The graph of \(y = \cos^{-1} x\) passes through the point \((-1, 0)\).
(C) The graph of \(y = \mathrm{cosec}^{-1} x\) passes through the point \(\left(-1, -\displaystyle\frac{\pi}{2}\right)\).
(D) The graph of \(y = \cot^{-1} x\) passes through the origin.
Which of the above statements are correct?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
A point \((a, b)\) lies on the graph of \(y = f(x)\) exactly when \(f(a) = b\). So for each statement we find the principal value at the given x.
\( \displaystyle \tan^{-1}(1) = \frac{\pi}{4}, \quad \cos^{-1}(-1) = \pi, \quad \mathrm{cosec}^{-1}(-1) = -\frac{\pi}{2}, \quad \cot^{-1}(0) = \frac{\pi}{2} \)
So the four graphs pass through \(\left(1, \frac{\pi}{4}\right)\), \((-1, \pi)\), \(\left(-1, -\frac{\pi}{2}\right)\) and \(\left(0, \frac{\pi}{2}\right)\).
Checking the statements
(A) The graph of \(y = \tan^{-1} x\) passes through \(\left(1, \frac{\pi}{4}\right)\). — Correct. \(\tan\frac{\pi}{4} = 1\) and \(\frac{\pi}{4}\) is in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
(B) The graph of \(y = \cos^{-1} x\) passes through \((-1, 0)\). — Incorrect. At \(x = -1\) the height is \(\cos^{-1}(-1) = \pi\); the height 0 is reached at \(x = 1\).
(C) The graph of \(y = \mathrm{cosec}^{-1} x\) passes through \(\left(-1, -\frac{\pi}{2}\right)\). — Correct. \(\mathrm{cosec}\left(-\frac{\pi}{2}\right) = \frac{1}{-1} = -1\), and \(-\frac{\pi}{2}\) is in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\).
(D) The graph of \(y = \cot^{-1} x\) passes through the origin. — Incorrect. At \(x = 0\) the height is \(\cot^{-1}(0) = \frac{\pi}{2}\); the value 0 is never taken, as the range is \((0, \pi)\).