Question

prove that:

fast and quick​

Answer 1

To prove ---> I think there is a mistake in question

1 - Sin θ

-------------- = tan² (π /4 - θ /2 )

1 + Sinθ

Proof --->

We have some formula

Sinθ = Cos (π /2 - θ )

Cos2A = 2 Cos² A - 1

Cos2A = 1 - 2 Sin² A

Applying these formulee

Now

1 - Sinθ = 1 - Cos ( π /2 - θ )

= 1 - { 1 - 2 Sin² ( π /4 - θ / 2 )}

= 1 - 1 + 2 Sin² ( π /4 - θ /2 )

= 2 Sin² (π /4 - θ/2)

1 + Sinθ = 1 + Cos ( π / 2 - θ )

= 1 + 2 Cos² ( π /4 - θ / 2 ) - 1

= 2 Cos² ( π /4 - θ / 2 )

1 - Sinθ

LHS = ----------------

1 + Sinθ

2 Sin²(π /4 - θ /2)

= --------------------------------

2 Cos²( π /4 - θ /2 )

Sin² (π /4 - θ / 2 )

= ---------------------------------

Cos² ( π / 4 - θ / 2 )

= tan² ( π / 4 - θ / 2 ) = RHS