Answer:
Step-by-step explanation:
[tex]sin^2\theta+cos^2\theta=1~=>sin\theta=\pm\sqrt{1-cos^2\theta}\\\\cos\theta=\frac{\sqrt{2}}{2}[/tex]
we'll took the negative solucion of sin because sin is negative in IV cuadrant
[tex]sin\theta=-\sqrt{1-(\frac{\sqrt{2}}{2})^2}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}[/tex]
[tex]tan\theta=\frac{\sin\theta}{\cos\theta}[/tex] => [tex]tan\theta=-1[/tex]
so, the right option is the last one
