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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2011 · Mathematics · Inverse Trigonometric Functions

What is the principal value of ^ -1 ( 2 3 )?

What is the principal value of $\sec ^{-1} \left(\frac{2}{\sqrt{3}}\right)?

Options

  1. Aπ2\frac{\pi}{2}
  2. Bπ3\frac{\pi}{3}
  3. Cπ4\frac{\pi}{4}
  4. Dπ6\frac{\pi}{6}

Correct answer

D. π6\frac{\pi}{6}

Step-by-step solution

sec⁡−1(23)=sec⁡−1(sec⁡30∘)=sec⁡−1(sec⁡π6)=π6.\begin{aligned} \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right) & =\sec ^{-1}\left(\sec 30^{{}^{\circ}}\right) \\ & =\sec ^{-1}\left(\sec \frac{\pi}{6}\right) \\ & =\frac{\pi}{6}.\end{aligned}
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