By the Pythagorean identity....
\(\sin^2\theta+\cos^2\theta\,=\,1\\~\\ \frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}\,=\,\frac{1}{\cos^2\theta}\\~\\ \tan^2\theta+1\,=\,\sec^2\theta\) Let's divide both sides of this equation by cos2θ, and simplify.
Now plug in -135 for sec θ and solve the equation for tan θ .
\( \tan^2\theta+1\,=\,(-135)^2\\~\\ \tan^2\theta+1\,=\,18225\\~\\ \tan^2\theta\,=\,18225-1\\~\\ \tan^2\theta\,=\,18224 \)
Since cot θ is positive, tan θ is positive. So take the positive square root of both sides.
\(\tan\theta\,=\,\sqrt{18224}\\~\\ \tan\theta\,=\,4\sqrt{1139}\)
And...
\(\frac{\sec\theta}{\tan\theta}\,=\,\sec\theta\div\tan\theta\,=\,\frac{1}{\cos\theta}\div\frac{\sin\theta}{\cos\theta}\,=\,\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}\,=\,\frac{1}{\sin\theta}\,=\,\csc\theta\)
So....
\(\csc\theta\,=\,\frac{\sec\theta}{\tan\theta}\,=\,\frac{-135}{4\sqrt{1139}}\,=\,\frac{-135\sqrt{1139}}{4556}\\~\\ \csc\theta\,=\,\frac{-135\sqrt{1139}}{4556}\)
.