For $0^{\circ}<\theta<90^{\circ}$, $\frac{1}{\cos \theta}+\frac{1}{\tan \theta-\sec \theta}$ is equal to:
Answer & explanation
Correct answer: option 3
\(\frac{1}{cosθ}\) + \(\frac{1}{tanθ - secθ}\)
= \(\frac{1}{cosθ}\) + \(\frac{1}{tanθ - secθ}\) × \(\frac{tanθ +secθ}{tanθ + secθ}\)
= \(\frac{1}{cosθ}\) + \(\frac{tanθ +secθ}{tan²θ - sec²θ}\)
{ We know , sec²θ - tan²θ = 1 }
= \(\frac{1}{cosθ}\) + \(\frac{tanθ +secθ}{ - 1 }\)
= secθ - tanθ - secθ
= - tanθ