Answer: \(\displaystyle y=\frac{1}{x-1}-2=\frac{3-2x}{x-1}\).

Answer: \(\displaystyle y=\frac{1}{x-1}-2=\frac{3-2x}{x-1}\).

Use the identity

\(\operatorname{sec}^{2}\theta=1+\tan^{2}\theta.\)

Since \(x=\operatorname{sec}^{2}\theta\),

\(\tan^{2}\theta=x-1.\)

Also,

\(\operatorname{cot}^{2}\theta=\frac{1}{\tan^{2}\theta}.\)

So

\(y+2=\frac{1}{x-1}.\)

Therefore

\(y=\frac{1}{x-1}-2=\frac{3-2x}{x-1}.\)