Principal value branch

The principal value of \(\sin^{-1}x\) lies in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Remember the range in exams.

The principal value of \(\cos^{-1}x\) is in \(\left[0,\pi\right]\). This is a common exam fact.

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

The principal value of \(\cot^{-1}x\) is taken in \(\left(0,\pi\right)\). Remember that both endpoints are excluded.

Since \(\sin 0=0\) and \(0\) lies in the principal range of \(\sin^{-1}x\), which is \([-\frac{\pi}{2},\frac{\pi}{2}]\), we have \(\sin^{-1}0=0\). Note that although \(\sin\pi=0\), \(\pi\) is not in the principal range so arcsin does not return \(\pi\). Also \(\sin(\frac{\pi}{2})=1\) and \(\sin(-\frac{\pi}{2})=-1\), so those choices are incorrect. Exam tip: always recall the principal value range for \(\sin^{-1}x\) when evaluating inverse trig values.

We have \(\sin\left(-\frac{\pi}{2}\right)=-1\). Hence \(\sin^{-1}\left(-1\right)=-\frac{\pi}{2}\).

The principal value of arccos, \(\cos^{-1}x\), lies in \([0,\pi]\). Since \(\cos0=1\) and 0 is within this principal range, \(\cos^{-1}1=0\). Closest distractors fail: \(\frac{\pi}{2}\) gives \(\cos=0\), \(\pi\) gives \(\cos=-1\), and \(-\frac{\pi}{2}\) is outside the principal range, so none are correct. Exam tip: memorize that the principal value range of arccos is \([0,\pi]\).

The principal value range of \(\cos^{-1}x\) is \([0,\pi]\). Since \(\cos\left(\frac{\pi}{2}\right)=0\), we have \(\cos^{-1}0=\frac{\pi}{2}\). Options A and B are incorrect because \(\cos0=1\) and \(\cos\pi=-1\). Option D \(\(-\frac{\pi}{2}\)\) is a common confusion with \(\sin^{-1}\), but arccos values lie between 0 and \(\pi\). Exam tip: memorize the principal range \(0\le\cos^{-1}x\le\pi\).

Since \(\cos\pi=-1\) and the principal value (range) of \(\cos^{-1}x\) is \([0,\pi]\), we have \(\cos^{-1}(-1)=\pi\). The option \(-\pi\) is misleading because although \(\cos(-\pi)=-1\), arccos returns the principal value in \([0,\pi]\), not a negative angle. Options 0 and \(\frac{\pi}{2}\) are incorrect because \(\cos0=1\) and \(\cos(\frac{\pi}{2})=0\). Exam tip: always remember the range of arccos is \([0,\pi]\).

Since \(\tan 0=0\), the angle whose tangent is 0 is 0, so \(\tan^{-1}0=0\). Also remember the principal value range of \(\tan^{-1}x\) is \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\), so values like \(\pi\) or \(\tfrac{\pi}{2}\) are not valid principal outputs. The closest distractor \(\tfrac{\pi}{4}\) is wrong because \(\tan(\tfrac{\pi}{4})=1\), not 0. Exam tip: always use the definition \(\tan(\theta)\) and the principal range of \(\tan^{-1}\) to find inverse values quickly.

Reason: \(\tan\frac{\pi}{4}=1\), and the principal value of \(\tan^{-1}x\) lies in \((-\frac{\pi}{2},\frac{\pi}{2})\), so \(\tan^{-1}1=\frac{\pi}{4}\). Why other choices are wrong: \(0\) is wrong because \(\tan0=0\); \(\frac{\pi}{6}\) is wrong because \(\tan\frac{\pi}{6}=1/\sqrt{3}\), not 1; \(\frac{\pi}{2}\) is not valid as \(\tan\) is undefined there. Exam tip: Memorise standard tan values and remember arctan's principal range to pick the correct branch quickly.

Since \(\tan\left(-\frac{\pi}{4}\right)=-1\) and \(-\frac{\pi}{4}\) lies in the principal range, the answer is \(-\frac{\pi}{4}\).

Since \(\cot(\pi/4)=1\) and the principal value of \(\cot^{-1}x\) is taken in the interval \(0<\cot^{-1}x<\pi\), we get \(\cot^{-1}1=\frac{\pi}{4}\). The closest distractor \(\frac{\pi}{2}\) is incorrect because \(\cot(\pi/2)=0\). Exam tip: remember the principal-value range of arccot is \((0,\pi)\).

Since \(\sec\theta=\dfrac{1}{\cos\theta}\) and \(\sec 0=1\), the principal value of \(\sec^{-1}1\) is \(0\). The common principal range for \(\sec^{-1}x\) is \([0,\pi]\) excluding \(\pi/2\), so \(\sec^{-1}1=0\). Option \(\pi\) is incorrect because \(\sec\pi=-1\). Options \(\tfrac{\pi}{2}\) and \(-\tfrac{\pi}{2}\) are invalid since \(\sec\) is undefined where \(\cos\theta=0\). Exam tip: always check the principal range and where the original trig function is defined before selecting the inverse value.

If \(\csc y=1\) then \(\sin y=1\). The general solution of \(\sin y=1\) is \(y=\frac{\pi}{2}+2k\pi\) (\(k\in\mathbb{Z}\)). The principal value for \(cosec^{-1}x\) is usually taken in \([ -\frac{\pi}{2},\frac{\pi}{2}]\setminus\{0\}\), and the only solution in this interval is \(\frac{\pi}{2}\). Therefore \(cosec^{-1}1=\frac{\pi}{2}\). About distractors: \(0\) is not allowed (\(\csc\) undefined), \(-\frac{\pi}{2}\) gives \(\csc=-1\), and at \(\pi\) \(\sin\pi=0\) so \(\csc\) is undefined. Exam tip: convert \(\csc y=\) value to \(\sin y=\) value and then pick the principal value from the defined range.

The principal value of inverse cosecant, \(cosec^{-1}x\), is usually taken in \([-\tfrac{\pi}{2},\;\tfrac{\pi}{2}]\) excluding 0. Since \(\sin(-\tfrac{\pi}{2})=-1\), we have \(cosec(-\tfrac{\pi}{2})=-1\); hence \(cosec^{-1}(-1)=-\tfrac{\pi}{2}\). Note that \(\tfrac{3\pi}{2}\) also satisfies \(cosec= -1\) but it lies outside the principal range, so it is not the principal value. Exam tip: always use the principal-range convention for inverse trigonometric functions when answering numerical MCQs.

The angle \(\frac{\pi}{6}\) lies in the principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Hence the value is \(\frac{\pi}{6}\).

The angle \(\frac{2\pi}{3}\) lies in the principal range \(\left[0,\pi\right]\) of \(\cos^{-1}x\). Hence the value is \(\frac{2\pi}{3}\).

The principal-value rule for inverse tangent is essential: tan⁻¹(tan θ) equals θ only when θ lies in the principal range (−π/2, π/2). The angle π/6 lies inside this interval. Since tan(π/6) = 1/√3, applying the principal inverse tangent gives tan⁻¹(1/√3) = π/6. Therefore option A is correct. Although 5π/6 has the same tangent as π/6 because tangent is periodic, it is outside the principal range and cannot be returned by tan⁻¹. Option C has the opposite sign, while tan(π/3) = √3, not 1/√3, so option D is also incorrect. The inverse function is not unrestricted cancellation; its principal range must always be checked.

Since \(\sin\frac{\pi}{6}=\frac{1}{2}\), \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\).