Question

prove .......................

Answer 1

replacing theta by x

$\frac{ \sin(x) - \sqrt{1 + \sin(2x) } }{ \cos(x) - \sqrt{1 + \sin(2x) } } = \cot(x)$

cotx = Cosx/Sinx
using cross multiplication

$\sin^2(x) - \sin(x)\sqrt{1 + \sin(2x)}= \cos^2(x) - \cos(x)\sqrt{1 + \sin(2x)}$

$\sin^2(x) - \cos^2(x) = \sin(x)\sqrt{1 + \sin(2x)} - \cos(x)\sqrt{1 + \sin(2x)}$

$(\sin(x) - \cos(x))\times(\sin(x) + \cos(x)) = (\sin(x)-\cos(x) )(\sqrt{1 + \sin(2x)}$

cancelling sinx - cosx from both sides

$(\sin(x) + \cos(x)) = \sqrt{1 + \sin(2x)}$

squaring both sides

$(\sin(x) + \cos(x))^2 = (\sqrt{1 + \sin(2x)})^2$

$\sin^2(x) + \cos^2(x) + 2\sin(x) \cos(x) = 1 + \sin(2x)$

as we know that
$\sin^2(x) + \cos^2(x) = 1$
$2\sin(x) \cos(x) = \sin(2x)$
using these values
$1 + \sin(2x) = 1 + \sin(2x)$

hence
LHS = RHS

QED

Answer 2

Let us First take L.H.S. and make it equal to R.H.S.

From L.H.S.,

First solving Numerator,

Sinθ - √(1 + Sin2θ) = Sinθ - √(1 + 2 SinθCosθ)

= Sinθ - √(Sin²θ + Cos²θ + 2SinθCosθ)

= Sinθ - √(Sinθ + Cosθ)²

= Sinθ - (Sinθ + Cosθ)

= Sinθ - Sinθ - Cosθ

= -Cosθ

Taking Denominator,

Cosθ - √(1 + Sin2θ) = Cosθ - √(Sin²θ + Cos²θ + 2 SinθCosθ)

= Cosθ - √(Sinθ + Cosθ)²

= Cosθ - Sinθ - Cosθ

= -Sinθ

Now, L.H.S. = Numerator/Denominator = -Cosθ/-Sinθ

= Cotθ

= R.H.S.

Hence Proved.

Hope it helps.