​ Need Some Help

\(sin^2θ +cos^2θ =1\)

so \(sinθ =\sqrt{1-cos^2θ}=\sqrt{\frac{36}{36}-\frac{1}{36}}=\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{6}\)

so \(tanθ=\frac{sinθ}{cosθ}= \frac{{\frac{\sqrt{35}}{6}}}{\frac{-1}{6}}=-\sqrt{35}\)

 

Hope this helps!