Question
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{7\pi}{3}\Big)$
$\sec^{-1}\Big(\sec\frac{7\pi}{3}\Big)$
✓
Answer
We have
$\sec^{-1}\Big(\sec\frac{7\pi}{3}\Big)=\sec^{-1}\Big[\sec\Big(2\pi+\frac{\pi}{3}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{\pi}{3}\Big)\Big]$
$=\frac{\pi}{3}$
$\sec^{-1}\Big(\sec\frac{7\pi}{3}\Big)=\sec^{-1}\Big[\sec\Big(2\pi+\frac{\pi}{3}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{\pi}{3}\Big)\Big]$
$=\frac{\pi}{3}$
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