∣∣∣√1−sinθ1+sinθ+√1+sinθ1−sinθ∣∣∣=−2cosθ,whereπ2<θπ
LHS=∣∣∣√1−sinθ1+sinθ+√1+sinθ1−sinθ∣∣∣=∣∣∣(√1−sinθ)2+(√1+sinθ)2√(1+sinθ)(1−sinθ)∣∣∣=∣∣∣1−sinθ+1+sinθ√1−sin2θ∣∣∣
=∣∣2cosθ∣∣(∵1−sin2θ=cos2θ⇒√1−sin2θ=cosθ)
=−2cosθ(∵π2<θ<π⇒cosθ<0)
=RHS Hence Proved.