Question

SecA-tanA=(cosA/2-sinA/2)/(cosA/2+sinA/2)​

Answer 1

Explanation:

SecA - TanA

= 1/CosA - SinA/CosA

= 1 - SinA/CosA

We know that Sin2A = 2SinACosA and Cos2A = Cos²A - Sin²A

Thus SinA = Sin2(A/2) = 2Sin(A/2)CosA/2

CosA = Cos2(A/2) = Cos²A/2 - Sin²A/2

Now substituting the values back,

=> 1 - 2Sin(A/2)Cos(A/2) / Cos²(A/2) - Sin²(A/2)

// we know that Sin²θ + Cos²θ = 1

=> Sin²(A/2) + Cos²A/2 - 2Sin(A/2)Cos(A/2) / Cos²(A/2) - Sin²(A/2)

//We know that numerator is of form a² + b² - 2ab which is (a - b)².

//Similarly denominator is of form a² - b² which is (a - b)(a + b)

=> [Sin(A/2) - Cos(A/2)]² / [Cos(A/2) + Sin(A/2)][Cos(A/2) - Sin(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)]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)]

= R.H.S

Hence proved.