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Model Paper 10 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 101 Mark
MCQ
Find the value of $\sec \left(\frac{-19 \pi}{3}\right)$.
  • A
    $\frac{1}{2}$
  • B
    -2
  • C
    2
  • D
    $\frac{-1}{2}$
✓

Answer

(c) 2
Explanation: $\sec \left(\frac{-19 \pi}{3}\right)=\sec \frac{19 \pi}{3}[\because \sec (-\theta)=\sec \theta]$
$=\sec \left(6 \pi+\frac{\pi}{3}\right)=\sec \frac{\pi}{3}=2[\because \sec (2 n \pi+\theta)=\sec \theta]$

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