Introduction to inverse trigonometric functions

The domain of \(\sec^{-1}x\) is \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\). The number (0) is not in this domain.

As (x\to\infty), (\tan^{-1}x\to\frac{\pi}{2}). The value (\frac{\pi}{2}) is a limit, not an attained value.

The principal range of tan⁻¹x is (−π/2, π/2). As x becomes increasingly negative, the angle in this range whose tangent equals x approaches the lower endpoint −π/2. In limit notation, lim x→−∞ tan⁻¹x = −π/2. Thus option A is correct. The value is approached from above; it is a horizontal asymptotic value rather than a value attained at a finite input. Option B is the corresponding limit when x→+∞, so its sign is wrong here. Option C would describe behavior near x = 0, not at negative infinity, and option D lies outside the principal range of arctangent. The sign and the direction of infinity are both essential in identifying the correct limit.

The value of (\sin x) lies only in ([-1,1]) so the domain of (\sin^{-1}x) is ([-1,1]). In exams check the domain first.

The value of (\cos x) also lies in ([-1,1]) so the domain of \(\cos^{-1}x\) is ([-1,1]). Check the range of (x) before substitution.

The principal value of \(\tan^{-1}x\) lies in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). The endpoints \(\pm\frac{\pi}{2}\) are not included.

Because (\cos 0=1) and (0\in[0,\pi]) so (\cos^{-1}1=0). For (\cos^{-1}x) the answer is not negative.

Because (\tan \frac{\pi}{4}=1) and (\frac{\pi}{4}) lies in the principal range. Remembering basic values helps solve quickly.

The domain of (\tan^{-1}x) is (\mathbb{R}) and (\tan(\tan^{-1}x)=x). This is a very common identity.

The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) so \(\sin^{-1}(\sin x)=x\) is directly true there. Without range it can be a mistake.

The principal range of \(\cos^{-1}x\) is \([0,\pi]\). Therefore \(\cos^{-1}(\cos x)=x\) applies directly only in this range.

The principal range of \(\tan^{-1}x\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Therefore that interval is the correct choice.

For every (x\in[-1,1]), (\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}). This identity is often used directly.

Taking (\sin) on both sides gives (x=\sin\frac{\pi}{6}=\frac{1}{2}). In inverse functions convert the angle back to the original value.

Taking (\cos) on both sides gives (x=\cos\frac{\pi}{3}=\frac{1}{2}). The angle lies in the principal range so the answer is valid.

Taking (\tan) on both sides gives (x=\tan\frac{\pi}{4}=1). In (\tan^{-1}) questions the value comes directly from tangent.

Since (\cot x) can take all real values the domain of (\cot^{-1}x) is (\mathbb{R}). Do not mix it with the domain of (\sin^{-1}x).

The principal value of \(\cot^{-1}x\) is usually taken in \((0,\pi)\). Hence \(\cot^{-1}0=\frac{\pi}{2}\).

The expression \(\sin^{-1}x\) is defined only for \(x\in[-1,1]\), and \(2\notin[-1,1]\). In domain questions check each option quickly.

For a real number y, sin y always lies between −1 and 1 inclusive. Therefore the inverse sine equation sin y = x has a real solution only when x belongs to the interval [−1, 1]. The endpoints are included because sin(−π/2) = −1 and sin(π/2) = 1. Hence the required condition is −1 ≤ x ≤ 1, making option A correct. Values greater than 1 or less than −1 cannot be outputs of the real sine function, so options B and C are impossible. Option D is false because inverse sine is not real for every real input; for example, sin⁻¹2 has no real value. This interval is the real domain of the principal inverse sine function.