Question

Prove the following: cosA / 1 - sinA = tan (45 + A/2)

Answer 1

LHS
= cosA ÷ (1 - sinA)
= [cosA ÷ (1 - sinA)] * [ (1 + sinA)÷(1 + sinA) ]   
= [cosA * (1 + sinA)] ÷ [(1 - sinA) * (1 + sinA) ]
= [cosA * (1 + sinA)] ÷ (1-sin²A)
= [cosA * (1 + sinA)] ÷ cos² A
= (1 + sinA)÷ cosA
= [ cos²(A/2) + sin²(A/2)+ 2sin(A/2)cos(A/2) ] ÷ [cos²(A/2) - sin²(A/2)]
= [ cos(A/2) + sin(A/2) ]² ÷ [ {cos(A/2) - sin(A/2)} * {cos(A/2) + sin(A/2)} ]
= [ cos(A/2) + sin(A/2) ] ÷ [cos(A/2) - sin(A/2)]
= { [ cos(A/2) + sin(A/2) ] ÷ [cos(A/2) - sin(A/2)] } * { [1÷ cos(A/2)] ÷[1÷ cos(A/2)] }
= [ 1+ tan(A/2) ] ÷ [ 1-tan(A/2) ] 
= [ tan 45° + tan(A/2) ] ÷ [ 1 - tan 45° tan(A/2) ]
= tan (45° + A/2) = RHS (hence proved)

Answer 2

Answer:

1-tan^2A/2

cosA = -----------------

1+tan^2A/2

2tanA/2

sinA = ----------------

1+ tan^2A/2

cosA

=----------

1-sinA

1-tan^2A/2

------------------

1+tan°2A/2

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

2tanA/2

1- -----------------

1+ tan^2A/2

1- tana^2A/2

---------------------

1+tan^2A

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

1+tan^2A/2 -2tanA/2

-----------------------------------

1+tan^2A/2

(1-tanA/2)(1+tanA/2)

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

( 1-tanA/2)^2

1+ tanA/2

=---------------. {tana= 1 ,tana= 45°}

1 - tanA/2

(45° + tanA/2)