For $0^{\circ}<\theta<90^{\circ}$, $\frac{1}{\cos \theta}+\frac{1}{\tan \theta-\sec \theta}$ is equal to:

-tan θ

\(\frac{1}{cosθ}\) + \(\frac{1}{tanθ - secθ}\)

= \(\frac{1}{cosθ}\) + \(\frac{1}{tanθ - secθ}\) × \(\frac{tanθ +secθ}{tanθ + secθ}\)

= \(\frac{1}{cosθ}\) + \(\frac{tanθ +secθ}{tan²θ - sec²θ}\)

{ We know ,  sec²θ - tan²θ = 1 }

= \(\frac{1}{cosθ}\) + \(\frac{tanθ +secθ}{ - 1 }\)

= secθ - tanθ - secθ

= - tanθ