Which of the following statements are correct?
(A) \(\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\displaystyle\frac{x+y}{1-xy}\right),\ xy \lt 1\)
(B) \(\tan^{-1}x+\tan^{-1}y = \pi + \tan^{-1}\left(\displaystyle\frac{x+y}{1-xy}\right),\ xy \gt 1\)
(C) \(\sin^{-1}x+\cos^{-1}x=\displaystyle\frac{\pi}{2},\ x \in (-1, 1)\)
(D) \(\tan^{-1}x+\cot^{-1}x=\displaystyle\frac{\pi}{2},\ x \in R\)
(E) \(\cos^{-1}(-x) = -\cos^{-1}x,\ x \in [-1, 1]\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C), (D) Only
The standard results on inverse trigonometric functions are:
\( \displaystyle \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy},\quad xy \lt 1 \)
\( \displaystyle \tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\frac{x+y}{1-xy},\quad xy \gt 1,\ x \gt 0,\ y \gt 0 \)
\( \displaystyle \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\ x \in [-1, 1];\quad \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2},\ x \in R \)
\(\cos^{-1}(-x) = \pi - \cos^{-1}x,\ x \in [-1, 1]\)
So (A), (C) and (D) are standard results, the π-form in (B) needs x, y both positive, and the correct form of (E) is \(\pi - \cos^{-1}x\).
Checking the statements
(A) sum formula for xy < 1 — Correct. It is the standard result.
(B) π + … for xy > 1 — Incorrect. For x = y = −2 (xy = 4) the left side \(2\tan^{-1}(-2)\) is negative, while the right side \(\pi + \tan^{-1}\frac{4}{3}\) is more than π.
(C) \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) for x in (−1, 1) — Correct. It holds on all of [−1, 1], so on (−1, 1) too.
(D) \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) for all real x — Correct. It is the standard result.
(E) \(\cos^{-1}(-x) = -\cos^{-1}x\) — Incorrect. At x = 0 the left side is \(\frac{\pi}{2}\) and the right side is \(-\frac{\pi}{2}\); the range of \(\cos^{-1}\) is [0, π], so it is never negative.
Note: This topic was not in the CUET 2026 syllabus.