Principal value branch

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 \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

Let \(\theta=\sec^{-1}(2)\). Then \(\sec\theta=2\), so \(\cos\theta=\frac{1}{2}\). In the principal range \([0,\pi]\), \(\cos\theta=\frac{1}{2}\) occurs at \(\theta=\frac{\pi}{3}\). Hence \(\sec^{-1}(2)=\frac{\pi}{3}\).

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, \(\cos\theta=\frac{12}{13}\).

Here \(\cos\theta=\frac{12}{13}\) and \(\theta\in\left[0,\pi\right]\). In this case, \(\sin\theta=\frac{5}{13}\) is positive.

Direct answer: option A, \(-\frac\pi4\). First evaluate the inner value: \(\sin(-\frac\pi4)=-\frac{\sqrt2}{2}\). Now inverse sine asks for the principal angle having that sine. Because \(-\frac\pi4\) already lies in the principal range of \(\sin^{-1}x\), namely \([-\frac\pi2,\frac\pi2]\), the outer function returns the same angle: \(\sin^{-1}(\sin(-\frac\pi4))=-\frac\pi4\). A is correct. B changes the sign and has positive sine. C, \(-\frac{3\pi}4\), is outside the principal range. D is positive and also outside that range. The inverse function does not always cancel sine, but it does here because the original angle is in the required range. Memory cue: for sine, keep angles between -90° and 90° unchanged.

Since \(\cos0=1\) and \(\cos^{-1}1\) is the angle in the principal range \([0,\pi]\) whose cosine equals 1, we get \(\cos^{-1}1=0\). The other choices are incorrect: \(\cos(\pi/2)=0\), \(\cos\pi=-1\), and \(-\pi\) is outside the principal range of arccos. Exam tip: When evaluating \(\cos^{-1}(\cos\theta)\), reduce or map \(\theta\) into the principal range \([0,\pi]\) and use that equivalent angle.