Consider the following statements:
(A) The equation \(\cos^{-1} x = -\displaystyle\frac{\pi}{3}\) has no real solution.
(B) The equation \(\tan^{-1} x = \displaystyle\frac{\pi}{2}\) has no real solution.
(C) The equation \(\sin^{-1} x = \displaystyle\frac{2\pi}{3}\) has the solution \(x = \displaystyle\frac{\sqrt{3}}{2}\).
(D) The equation \(\cot^{-1} x = \displaystyle\frac{\pi}{3}\) has the solution \(x = \displaystyle\frac{1}{\sqrt{3}}\).
Which of the above statements are correct?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
An equation of the form (inverse function of x) = \(\alpha\) has a solution only when \(\alpha\) lies in the range of the principal value branch; the solution is then the trigonometric value of \(\alpha\).
\(\cos^{-1}: [0, \pi]\)
\( \displaystyle \tan^{-1}: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad \sin^{-1}: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
\(\cot^{-1}: (0, \pi)\)
So \(-\frac{\pi}{3}\), \(\frac{\pi}{2}\) and \(\frac{2\pi}{3}\) are outside the ranges they are set against, while \(\frac{\pi}{3}\) is inside \((0, \pi)\).
Checking the statements
(A) \(\cos^{-1} x = -\frac{\pi}{3}\) has no real solution. — Correct. Values of \(\cos^{-1}\) lie in \([0, \pi]\) and are never negative.
(B) \(\tan^{-1} x = \frac{\pi}{2}\) has no real solution. — Correct. \(\frac{\pi}{2}\) is an open end of \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) and is never taken.
(C) \(\sin^{-1} x = \frac{2\pi}{3}\) has the solution \(x = \frac{\sqrt{3}}{2}\). — Incorrect. \(\frac{2\pi}{3}\) is greater than \(\frac{\pi}{2}\), so there is no solution; in fact \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\).
(D) \(\cot^{-1} x = \frac{\pi}{3}\) has the solution \(x = \frac{1}{\sqrt{3}}\). — Correct. \(\frac{\pi}{3}\) lies in \((0, \pi)\) and \(\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}\).