Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINVERSE TRIGNOMETRIC FUNCTIONS2 Marks
Question
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)$
✓
Answer
We have
$\sec^{-1}\Big(\sec\frac{13\pi}{4}\Big)=\sec^{-1}\Big[\sec\Big(4\pi-\frac{3\pi}{4}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{3\pi}{4}\Big)\Big]$
$=\frac{3\pi}{4}$
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