Question

tan(45°-A/2)=1-sin A/cos A





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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)

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