What is the value of $\frac{tanθ - secθ + 1}{tanθ+sec θ - 1}$ ?
Answer & explanation
Correct answer: option 4
\(\frac{tanθ - secθ + 1}{ tanθ + secθ - 1 }\)
{ we know, sec²θ - tan²θ = 1 }
= \(\frac{tanθ - secθ + 1}{ tanθ + secθ - sec²θ - tan²θ }\)
= \(\frac{tanθ - secθ + 1 }{ tanθ + secθ - (secθ - tanθ) .(secθ + tanθ)}\)
= \(\frac{tanθ - secθ + 1}{ tanθ + secθ - (secθ - tanθ) .(secθ + tanθ) }\)
= \(\frac{tanθ - secθ + 1}{ tanθ + secθ[1 - (secθ - tanθ)] }\)
= \(\frac{1 }{ tanθ + secθ}\)
= \(\frac{cosθ }{ 1+sinθ}\)