Question

Solve it plz. I want it fast

Answer 1

pls refer the picture for answrr

Answer 2

Answer:

$secθ + tanθ = p \\ secθ - tanθ = \frac{1}{p} \\ secθ - tanθ = \frac{1}{secθ + tanθ } \\ cross \: multiply \\ (secθ - tanθ)(secθ + tanθ)= 1 \\ {(secθ)}^{2} - {(tanθ)}^{2} = 1 \\ {sec}^{2} θ - {tan}^{2} θ = 1 \\ 1 = 1 \\ LHS=RHS \\ Hence \: proved$