What is the principal value of \(\cos^{-1}(0)\)?
Correct answer: A
\(\cos\frac{\pi}{2}=0\) and \(\frac{\pi}{2}\in[0,\pi]\). Remember special values along with the principal range.
View question detailsMathematics
Principal value branch
प्रधान मान शाखा
In this Class 12 Mathematics topic from Inverse Trigonometric Functions, students learn why trigonometric functions must be restricted to suitable domains to have well-defined inverses. The topic explains the principal value branch, standard ranges of sin⁻¹x, cos⁻¹x, tan⁻¹x and related functions, and how these ranges determine the unique value returned by an inverse function. Students also practise interpreting graphs, selecting correct branches, and applying principal-value conventions while simplifying expressions and solving problems.
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Medium · Level 32 · inverse-trigonometric-functions,special-value,cos-inverse
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- (\frac{\pi}{2})
- (0)
- (\pi)
- -(\frac{\pi}{2})
Medium · Level 32 · inverse-trigonometric-functions,special-values,principal-value,tan-inverse,arctan
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- 0
- \frac{\pi}{2}
- \pi
- -\frac{\pi}{2}
Medium · Level 32 · inverse-trigonometric-functions,trigonometry,arccos,pythagorean-identity,quadrants
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- \(\dfrac{4}{5}\)
- \(-\dfrac{4}{5}\)
- \(\dfrac{3}{5}\)
- \(-\dfrac{3}{5}\)
Medium · Level 32 · inverse-trigonometric-functions,evaluation,sum
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- \(\frac{\pi}{3}\)
- \(\frac{\pi}{6}\)
- \(\frac{\pi}{2}\)
- \(\frac{2\pi}{3}\)
Medium · Level 32 · inverse-trigonometric-functions,special-values,principal-values,trigonometry
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- 0
- \(\frac{\pi}{4}\)
- \(-\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
Medium · Level 32 · inverse-trigonometric-functions,principal-value,sin-composition
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- -\(\frac{2\pi}{3}\)
- -\(\frac{\pi}{3}\)
- \(\frac{\pi}{3}\)
- \(\frac{2\pi}{3}\)
Medium · Level 32 · inverse-trigonometric-functions,principal-value,cos-composition
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- -\(\frac{\pi}{3}\)
- \(\frac{\pi}{3}\)
- \(\frac{2\pi}{3}\)
- \(\frac{4\pi}{3}\)
Medium · Level 32 · inverse-trigonometric-functions,principal-value,tan-composition
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- -(\frac{5\pi}{6})
- (\frac{\pi}{6})
- -(\frac{\pi}{6})
- (\frac{5\pi}{6})
Medium · Level 32 · inverse-trigonometric-functions,triangle-method,negative-input
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- \(\frac{12}{13}\)
- -\(\frac{12}{13}\)
- \(\frac{5}{13}\)
- -\(\frac{5}{13}\)
Medium · Level 32 · inverse trigonometric functions, principal value range, arccos, class 12 mathematics, trigonometry
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- \(\sin^{-1}x\)
- \(\cos^{-1}x\)
- \(\tan^{-1}x\)
- \(\cot^{-1}x\)
What is the principal value of \((\tan^{-1}0)\)?
Correct answer: A
Since \(\tan0=0\) and the principal range of \(\tan^{-1}x\) is \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\), the principal value is \(0\). The values \(\pm\tfrac{\pi}{2}\) are not valid principal values (\tan\) is undefined there, and \(\pi\) lies outside the principal interval. Exam tip: always verify the principal range of the inverse trig function before selecting the value.
View question detailsIf \(\theta=\cos^{-1}\left(-\dfrac{3}{5}\right)\), what is the value of \(\sin\theta\)?
Correct answer: A
From \(\cos\theta=-\dfrac{3}{5}\) and \(\sin^2\theta+\cos^2\theta=1\) we get \(\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\), so \(\sin\theta=\pm\dfrac{4}{5}\). Since \(\theta=\cos^{-1}(\ldots)\) lies in \([0,\pi]\) and \(\cos\theta<0\) places \(\theta\) in the second quadrant where sine is positive, \(\sin\theta=\dfrac{4}{5}\). Closest trap is \(-\dfrac{4}{5}\) from ignoring the quadrant sign. Exam tip: always use the principal range of inverse cosine and the identity \(\sin^2+\cos^2=1\).
View question detailsWhat is the value of \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sin^{-1}\left(\frac{1}{2}\right)\)?
Correct answer: A
Here \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\) and \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). Hence the sum is \(\frac{\pi}{3}\).
View question detailsFind the value of \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)-\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)\).
Correct answer: A
The principal value range of \(\sin^{-1}x\) is \([-\tfrac{\pi}{2},\tfrac{\pi}{2}])\) and for \(\cos^{-1}x\) it is \([0,\pi]\). Since \(\sin\tfrac{\pi}{4}=\cos\tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}\), both inverse values equal \(\tfrac{\pi}{4}\). Therefore the difference is \(\tfrac{\pi}{4}-\tfrac{\pi}{4}=0\). The nearest distractor \(\tfrac{\pi}{4}\) is the value of each inverse separately, not their difference. Exam tip: always check the principal value intervals for inverse trig functions before subtracting.
View question detailsWhat is the principal value of \(\sin^{-1}\left(\sin\left(-\frac{2\pi}{3}\right)\right)\)?
Correct answer: B
\(\sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\), so the principal value is -\(\frac{\pi}{3}\). Keep the answer of \(\sin^{-1}x\) in \(-[\frac{\pi}{2},\frac{\pi}{2}]\).
View question detailsWhat is the principal value of \(\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)\)?
Correct answer: B
\(\cos\left(-\frac{\pi}{3}\right)=\frac{1}{2}\) and \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\). The answer of \(\cos^{-1}x\) is not negative.
View question detailsWhat is the principal value of \(\tan^{-1}\left(\tan\left(-\frac{5\pi}{6}\right)\right)\)?
Correct answer: B
\(\(\tan\left(-\frac{5\pi}{6}\right)=\frac{1}{\sqrt{3}}\), so the principal value is \(\frac{\pi}{6}\). Give the answer of \(\tan^{-1}x\) in ((-\frac{\pi}{2},\frac{\pi}{2})\).
View question detailsWhat is the value of \(\cos\left(\sin^{-1}\left(-\frac{5}{13}\right)\right)\)?
Correct answer: A
The principal angle of \(\sin^{-1}\) may lie in the fourth quadrant, where \(\cos\theta\) is positive. From the (5,12,13) triangle, the answer is \(\frac{12}{13}\).
View question detailsWhich of the following inverse trigonometric functions has the principal-value range \([0,\pi]\)?
Correct answer: B
The principal-value range of \(\cos^{-1}x\) is \([0,\pi]\), where \(x\in[-1,1]\). Hence, option B is correct. The closest distractor, \(\sin^{-1}x\), has principal-value range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so it is not correct. Exam tip: Memorise the principal-value ranges of inverse trigonometric functions separately.
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