Introduction to inverse trigonometric functions

Correct answer: A. \(\frac{\pi}{6}\)

Because \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\). Inverse values are found quickly from basic trigonometric values.

Correct answer: A. \(\frac{\pi}{4}\)

Since \(\frac{\pi}{4}\in[0,\pi]\), \(\cos^{-1}\left(\cos\frac{\pi}{4}\right)=\frac{\pi}{4}\). If the angle lies in the principal range, it remains unchanged.

Correct answer: A. \(\frac{\pi}{3}\)

The angle \(\frac{\pi}{3}\) lies in the principal range \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Therefore the value of the expression is \(\frac{\pi}{3}\).

Correct answer: A. (\frac{\pi}{6})

Because (\operatorname{cosec}\frac{\pi}{6}=2). In (\operatorname{cosec}^{-1}x), the magnitude of (x) cannot be less than (1).

Correct answer: A. (\frac{\pi}{6})

Because (\cot\frac{\pi}{6}=\sqrt{3}) and (\frac{\pi}{6}\in(0,\pi)). Remember ((0,\pi)) for the principal value of (\cot^{-1}).

Correct answer: A. \(\cos^{-1}\left(\frac{7}{6}\right)\)

The expression \(\cos^{-1}x\) is real only for \(x\in[-1,1]\), and \(\frac{7}{6}>1\). Hence it will not give a real value.

Correct answer: A. \(\frac{2\pi}{3}\)

The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(\frac{2\pi}{3}\) lies outside this range.

Correct answer: A. (-\frac{\pi}{3})

The range of (\cos^{-1}x) is ([0,\pi]), so a negative value is not possible. In such questions check only the output range.

Correct answer: A. \(\frac{\pi}{2}\)

The range of \(\tan^{-1}x\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), which does not include \(\frac{\pi}{2}\). Pay attention to open intervals.

Correct answer: A. \(\frac{4}{5}\)

Let \(\theta=\cos^{-1}\frac{3}{5}\), then \(\cos\theta=\frac{3}{5}\) and \(\theta\in[0,\pi]\). Here \(\sin\theta=\frac{4}{5}\).

Correct answer: A. \(\frac{\pi}{4}\)

Because \(\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\). Standard angle values are very useful in exams.

Correct answer: A. \(\frac{3\pi}{4}\)

Because \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) and \(\frac{3\pi}{4}\in[0,\pi]\). For \(\cos^{-1}\), do not take a negative principal value.

Correct answer: A. \(\frac{12}{13}\)

Let \(\theta=\sin^{-1}\frac{5}{13}\), then \(\sin\theta=\frac{5}{13}\) and \(\theta\) is in the principal range. Therefore \(\cos\theta=\frac{12}{13}\).

Correct answer: A. \(\frac{3}{4}\)

Let \(\theta=\cos^{-1}\frac{4}{5}\), then \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\). Hence \(\tan\theta=\frac{3}{4}\).

Correct answer: A. \(\left[-\frac{1}{2},\frac{1}{2}\right]\)

For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).

Correct answer: A. ([-3,3])

For \(\cos^{-1}\left(\frac{x}{3}\right)\), we need \(-1\le\frac{x}{3}\le1\). Therefore \(-3\le x\le3\).

Correct answer: A. (\frac{\pi}{2})

When (x>0), (\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}). Keep the principal range of (\cot^{-1}x) in mind.

Correct answer: A. \(-\frac{\pi}{6}\)

Because \(\operatorname{cosec}\left(-\frac{\pi}{6}\right)=-2\) and \(-\frac{\pi}{6}\) lies in the principal range. For negative values choose the correct principal angle.

Correct answer: A. \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

The value of \(\sin^{-1}x\) always lies in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), where \(\sin x\) is one-one. The interval \([0,\pi]\) is the principal value range of \(\cos^{-1}x\), not inverse sine. Exam tip: memorise the standard principal ranges of inverse trigonometric functions.