Introduction to inverse trigonometric functions

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.

Let \(y=\tan^{-1}x\). Then by definition, \(\tan y=x\) (with \(y\) in the principal range). Therefore \(\tan(\tan^{-1}x)=\tan y=x\).

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.

If (\cos^{-1}x=0), then (x=\cos0=1). In an inverse trigonometric equation apply the corresponding trigonometric function on both sides.

Taking (\sin) on both sides gives (x=\sin\frac{\pi}{2}=1). This is the maximum principal value of (\sin^{-1}x).

Since \(\sin(5\pi/6)=1/2\), we need the angle in the principal range \([-\pi/2,\pi/2]\) whose sine is \(1/2\). It is \(\pi/6\), not \(5\pi/6\). Exam tip: always check the principal-value range first.

Since (\sin^{-1}(-x)=-\sin^{-1}x), (\sin^{-1}x) is odd. The function (\tan^{-1}x) is also odd but is not in the options.

The function (\sin^{-1}x) is odd, so (\sin^{-1}(-x)=-\sin^{-1}x). This identity is useful in sign-change questions.

The direct answer is A: \(\tan^{-1}(-x)=-\tan^{-1}x\). The symbol \(\tan^{-1}\) here means the inverse tangent function, not the reciprocal. A function is called odd when changing its input from \(x\) to \(-x\) changes its output to its negative. In general, if \(y=\tan^{-1}x\), then \(\tan y=x\), where the principal value of \(y\) lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). Since tangent is an odd function, \(\tan(-y)=-\tan y=-x\). Therefore the principal inverse-tangent value for \(-x\) is \(-y=-\tan^{-1}x\). Option A is correct because it states this exact identity. Option B, \(\pi-\tan^{-1}x\), is not generally in the principal range and does not give the inverse tangent of a negative input. Option C uses a shift by \(\frac{\pi}{2}\), which has no valid basis here. Option D gives \(\tan^{-1}x\), ignoring the sign change, so it is wrong except in the special case \(x=0\). Memory cue: inverse tangent is odd, so simply put a minus sign outside.

The angle \(-\frac{\pi}{6}\) lies in the principal range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), so the value is the same. For \(\tan^{-1}\), the interval is open.

The statement \(y=\sin^{-1}x\) means \(\sin y=x\) and (y) lies in the principal range. Understanding inverse functions by definition is the safest method.

The expression \(y=\tan^{-1}x\) means \(\tan y=x\). Here (y) stays in the principal range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

Because \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\frac{\pi}{3}\) lies in the principal range. For \(\tan^{-1}\), take the angle in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).