Question

Prove that cosA / 1-sinA = (1+ tanA/2) / (1- tanA/2)

Answer 1

Answer:

Step-by-step explanation:

Formula used:

(1.) cos[2(a/2)]

= cos^2(a/2) - sin^2(a/2)

= [cos(a/2) - sin(a/2)]*[cos(a/2) + sin(a/2)]

(2.) 1 - sin[2(a/2)]

= cos^2(a/2) + sin^2(a/2) - 2.sin(a/2).cos(a/2)

= [cos(a/2) - sin(a/2)]^2

= [cos(a/2) - sin(a/2)]*[cos(a/2) - sin(a/2)]

Now substituting 1. in numerator and 2. in denominator we get

{[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)]

= Dividing numerator and denominator by cos(a/2) we get

[1 + tan(a/2)]÷[1 - tan(a/2)]

Answer 2

Answer:

= cos^2(a/2) - sin^2(a/2)

cos^2(a/2) + sin^2(a/2) - 2.sin(a/2).cos(a/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)]

={[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)]

= Dividing numerator and denominator by cos(a/2) we get

=[1 - tan(a/2)]    = R.H.S. proved

 [1 + tan(a/2)]

Step-by-step explanation: