Principal value branch

Since \(\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\) and the principal value of \(\sin^{-1}x\) lies in \([-\frac{\pi}{2},\frac{\pi}{2}]\), we have \(\sin^{-1}\!\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\). The option \(-\frac{\pi}{3}\) is incorrect because \(\sin(-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}\). \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) are wrong as their sines are \(\frac{1}{2}\) and \(\frac{\sqrt{2}}{2}\) respectively. Exam tip: memorize standard sine/cosine values for common angles \(0, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), \(\frac{\pi}{2}\)\).

Since \(\cos\frac{\pi}{3}=\frac{1}{2}\), \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\).

The principal value of arccos is taken in the interval \([0,\pi]\). Since \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\), we have \(\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}\). The other choices are incorrect: \(\cos\frac{\pi}{3}=\tfrac{1}{2},\;\cos\frac{\pi}{4}=\tfrac{\sqrt{2}}{2}\), and \(\cos\frac{5\pi}{6}=-\tfrac{\sqrt{3}}{2}\). Exam tip: memorize standard trig values and remember arccos range \([0,\pi]\).

Reason: \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\), and the principal value of arctan lies in \((-\frac{\pi}{2},\frac{\pi}{2})\). Hence \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\). Checking close distractors: \(\tan\frac{\pi}{4}=1\), so \(\frac{\pi}{4}\) is incorrect; \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\tan\frac{\pi}{2}\) is undefined. Exam tip: Memorise standard angle values and remember the principal range for inverse trig functions when answering quickly.

Since \(\cos\frac{2\pi}{3}=-\frac{1}{2}\) and \(\frac{2\pi}{3}\) is in the principal range, the answer is \(\frac{2\pi}{3}\).

Since \(sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}\) and \(sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\). Always choose the principal range value.

We have \(tan\frac{3\pi}{4}=-1\) and \(tan^{-1}\left(-1\right)=-\frac{\pi}{4}\). Remember the principal range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

If \(\sec\theta=2\), then \(\cos\theta=\dfrac{1}{2}\) because \(\sec\theta=\dfrac{1}{\cos\theta}\). For the principal value, \(\sec^{-1}x\) is taken in \([0,\pi]\) with \(\theta\neq\tfrac{\pi}{2}\).\nSolving \(\cos\theta=\tfrac{1}{2}\) gives \(\theta=\tfrac{\pi}{3}\), which lies in the principal range. Hence \(\sec^{-1}2=\tfrac{\pi}{3}\).\nWhy distractors fail: at \(\tfrac{\pi}{6}\), \(\sec=2/\sqrt{3}\); at \(\tfrac{\pi}{4}\), \(\sec=\sqrt{2}\); at \(\tfrac{2\pi}{3}\), \(\sec=-2\).\nExam tip: convert \(\sec^{-1}x\) to \(\cos\theta=1/x\) and then choose the solution inside the principal-value interval.

Since \(cosec\,\frac{\pi}{6}=\dfrac{1}{\sin(\pi/6)}=\dfrac{1}{1/2}=2\), the principal value is \(cosec^{-1}2=\frac{\pi}{6}\). The other options are incorrect because \(cosec\,\frac{\pi}{3}=\dfrac{2}{\sqrt{3}}\), \(cosec\,\frac{\pi}{4}=\sqrt{2}\) and \(cosec\,\frac{\pi}{2}=1\), none of which equals 2. Exam tip: use \(cosec\theta=1/\sin\theta\) and remember standard sine values for common angles.

The relation \(cos^{-1}\left(-x\right)=\pi-\cos^{-1}x\) is correct because of principal values. Do not treat it as an odd function.

Since \(cot\frac{3\pi}{4}=-1\) and \(frac{3\pi}{4}\in\left(0,\pi\right)\), the principal value is \(frac{3\pi}{4}\).

We have \(cot\frac{5\pi}{6}=-\sqrt{3}\). The principal value of \(cot^{-1}x\) must be taken in \(\left(0,\pi\right)\).

Core idea: use the identity \(\sec^{-1}x=\cos^{-1}\frac{1}{x}\) with the principal branch \([0,\pi]\) excluding \(\pi/2\). Thus \(\sec^{-1}2=\cos^{-1}\tfrac{1}{2}\). Since \(\cos\tfrac{\pi}{3}=\tfrac{1}{2}\), we get \(\cos^{-1}\tfrac{1}{2}=\tfrac{\pi}{3}\). Hence the principal value is \(\tfrac{\pi}{3}\). Closest distractor \(\tfrac{2\pi}{3}\) is incorrect because \(\sec\tfrac{2\pi}{3}=-2\), not +2. Exam tip: convert \(\sec^{-1}x\) to \(\cos^{-1}(1/x)\) and use standard cosine values (\(\pi/6,\pi/3,\pi/2\), etc.).

Since \(sec\frac{2\pi}{3}=-2\) and it lies in the principal range of \(sec^{-1}x\), the answer is \(\frac{2\pi}{3}\).

The standard principal value range of \(\cot^{-1}x\) is \((0,\pi)\). On this open interval, \(\cot\theta\) gives each real value exactly once, so the inverse has a unique principal value. The interval \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) is the principal value range for \(\tan^{-1}x\), not for \(\cot^{-1}x\). Exam tip: Memorise the principal value ranges of inverse trigonometric functions separately.

Since \(sin\frac{5\pi}{6}=\frac{1}{2}\) and \(sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). Always choose the angle in the principal range.

The angle \(\frac{5\pi}{6}\) lies in the principal range \(\left[0,\pi\right]\). Hence \(cos^{-1}\left(cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}\).

The principal value of \(\tan^{-1}x\) lies in \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\). The angle \(-\tfrac{\pi}{3}\) is inside this interval and \(\tan(-\tfrac{\pi}{3})=-\sqrt{3}\). Therefore \(\tan^{-1}(\tan(-\tfrac{\pi}{3}))=\tan^{-1}(-\sqrt{3})=-\tfrac{\pi}{3}\). Option B has the wrong sign; options C and D correspond to angles outside the principal range. Exam tip: always check whether the original angle lies in the principal branch of the inverse trig function before simplifying.

We have \(tan\frac{5\pi}{6}=-\frac{1}{\sqrt{3}}\). Thus \(tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}\).

Here \(\theta\) is in the principal range and \(\sin\theta=\frac{5}{13}\). From a right triangle, \(\cos\theta=\frac{12}{13}\).