If \(\frac{secθ - tanθ}{secθ + tanθ}\) = \(\frac{4}{5}\), find sinθ .

1 0 \(\frac{9}{8}\) \(\frac{1}{9}\)

\(\frac{1}{9}\)

\(\frac{secθ - tanθ}{secθ + tanθ}\) = \(\frac{4}{5}\) By componendo & Dividendo concept: \(\frac{secθ}{tanθ}\) = \(\frac{5 + 4}{5 - 4}\) \(\frac{\frac{1}{cos}}{\frac{sin}{cos}}\) = \(\frac{9}{1}\) \(\frac{1}{sinθ}\) = \(\frac{9}{1}\) ⇒ sinθ = \(\frac{1}{9}\)   (where 1 → P and 9 → H)