Question

Sec^-1(-square root 2)

Answer 1

`\bf \textit{let's say that, is an angle }\theta\quad so\qquad sec^(-1)(-√(2))=\theta \\\\\\ \textit{this means}\qquad sec(\theta )=-√(2)\quad but\quad sec(\theta)=\cfrac{1}{cos(\theta)}\qquad then \\\\\\ \cfrac{1}{cos(\theta)}=-√(2)\implies \cfrac{1}{-√(2)}=cos(\theta ) \\\\\\ \textit{now, if we rationalize the denominator, we get }-\cfrac{√(2)}{2} \\\\\\ -\cfrac{√(2)}{2}=cos(\theta )\implies cos^(-1)\left( -(√(2))/(2) \right)=cos^(-1) [cos(\theta )]`


`\bf cos^(-1)\left( -(√(2))/(2) \right)=\measuredangle \theta \implies \measuredangle \theta = \begin{cases} (3\pi )/(4)\\ (5\pi )/(4) \end{cases}`