# Find the exact value of the trigonometric function sec ((-9pi)/4) .

Find the exact value of the trigonometric function $\mathrm{sec}\left(\frac{-9\pi }{4}\right)$ .
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izboknil3
We know that
$\mathrm{sec}\left(-x\right)=\mathrm{sec}\left(x\right)$
Therefore,
$\mathrm{sec}\left(\frac{-9\pi }{4}\right)=\mathrm{sec}\left(\frac{9\pi }{4}\right)$
Rewriting the above trigonometric function,
$\mathrm{sec}\left(\frac{9\pi }{4}\right)=\mathrm{sec}\left(\frac{8\pi +\pi }{4}\right)$
$=\mathrm{sec}\left(\frac{8\pi }{4}+\frac{\pi }{4}\right)$
$=\mathrm{sec}\left(2\pi +\frac{\pi }{4}\right)$
We know that $2\pi +\theta$ lies in first quadrant and in first quadrant all trigonometric functions are positive.
Therefore,
$\mathrm{sec}\left(2\pi +\frac{\pi }{4}\right)=\mathrm{sec}\left(\frac{\pi }{4}\right)$
$=\sqrt{2}$
Hence, exact value of given trigonometric function is $\sqrt{2}$
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