-1-sin-1-sin-1-sin-1-sin-2-cos- where -2-

LHS= | √(1 sin θ/1+sin θ)+√(1+sin θ/1 sin θ) | = | (√(1 sin θ))^2+(√(1+sin θ))^2/√((1+sin θ)(1 sin θ)) |= | 1 sin θ+1+sin θ/√(1 sin^2 θ) | = | 2/cos θ | ( ∵ 1 ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

|√1-sinθ/1+si...

Question

$∣ ∣ ∣ \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} + \sqrt{\frac{1 + s i n θ}{1 - s i n θ}} ∣ ∣ ∣ = - \frac{2}{c o s θ}, w h e r e \frac{π}{2} < θ π$

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Solution

$L H S = ∣ ∣ ∣ \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} + \sqrt{\frac{1 + s i n θ}{1 - s i n θ}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{(\sqrt{1 - s i n θ})^{2} + (\sqrt{1 + s i n θ})^{2}}{\sqrt{(1 + s i n θ) (1 - s i n θ)}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{1 - s i n θ + 1 + s i n θ}{\sqrt{1 - s i n^{2} θ}} ∣ ∣ ∣$

$= ∣ ∣ \frac{2}{c o s θ} ∣ ∣ (∵ 1 - s i n^{2} θ = c o s^{2} θ \Rightarrow \sqrt{1 - s i n^{2} θ} = c o s θ)$

$= \frac{- 2}{c o s θ} (∵ \frac{π}{2} < θ < π \Rightarrow c o s θ < 0)$

=RHS Hence Proved.

Suggest Corrections

∣∣∣√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.

$∣ ∣ ∣ \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} + \sqrt{\frac{1 + s i n θ}{1 - s i n θ}} ∣ ∣ ∣ = - \frac{2}{c o s θ}, w h e r e \frac{π}{2} < θ π$