Question

Please help I will give brainliest!!!!!

Answer 1

Answer:

Step-by-step explanation:

$sin^2\theta+cos^2\theta=1~=>sin\theta=\pm\sqrt{1-cos^2\theta}\\\\cos\theta=\frac{\sqrt{2}}{2}$

we'll took the negative solucion of sin because sin is negative in IV cuadrant

$sin\theta=-\sqrt{1-(\frac{\sqrt{2}}{2})^2}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$

$tan\theta=\frac{\sin\theta}{\cos\theta}$  => $tan\theta=-1$

so, the right option is the last one