[tex]\bf \textit{let's say that, is an angle }\theta\quad so\qquad sec^{-1}(-\sqrt{2})=\theta
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\textit{this means}\qquad sec(\theta )=-\sqrt{2}\quad but\quad sec(\theta)=\cfrac{1}{cos(\theta)}\qquad then
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\cfrac{1}{cos(\theta)}=-\sqrt{2}\implies \cfrac{1}{-\sqrt{2}}=cos(\theta )
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\textit{now, if we rationalize the denominator, we get }-\cfrac{\sqrt{2}}{2}
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-\cfrac{\sqrt{2}}{2}=cos(\theta )\implies cos^{-1}\left( -\frac{\sqrt{2}}{2} \right)=cos^{-1} [cos(\theta )][/tex]
[tex]\bf cos^{-1}\left( -\frac{\sqrt{2}}{2} \right)=\measuredangle \theta \implies \measuredangle \theta =
\begin{cases}
\frac{3\pi }{4}\\
\frac{5\pi }{4}
\end{cases}[/tex]
