CUET Preparation Today
What is the value of $\frac{tanθ - secθ + 1}{tanθ+sec θ - 1}$ ? |
sec θ + tanθ $\frac{(1+sinθ)}{secθ}$ 2sec θ $\frac{cosθ}{(1+sinθ)}$ |
$\frac{cosθ}{(1+sinθ)}$ |
\(\frac{tanθ - secθ + 1}{ tanθ + secθ - 1 }\) { we know, sec²θ - tan²θ = 1 } = \(\frac{tanθ - secθ + 1}{ tanθ + secθ - sec²θ - tan²θ }\) = \(\frac{tanθ - secθ + 1 }{ tanθ + secθ - (secθ - tanθ) .(secθ + tanθ)}\) = \(\frac{tanθ - secθ + 1}{ tanθ + secθ - (secθ - tanθ) .(secθ + tanθ) }\) = \(\frac{tanθ - secθ + 1}{ tanθ + secθ[1 - (secθ - tanθ)] }\) = \(\frac{1 }{ tanθ + secθ}\) = \(\frac{cosθ }{ 1+sinθ}\) |
