prove: tan inverse [ cos x / (1+sin x)] = pie /4 - (x/2)
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= cos2(x/2) - sin2 (x/2) / sin2(x/2)+cos2(x/2)+2 sin(x/2) cos(x/2)
= [cos(x/2) - sin(x/2)] [cos(x/2) + sin(x/2)] / [sin(x/2)+cos(x/2)]2
= [cos(x/2) - sin(x/2)] / [cos(x/2) + sin (x/2)]
= 1 - tan(x/2) / 1+tan(x/2)
= tan (pi/4 - x/2)
=tan-1[tan(pi/4 - x/2)]
= pi/4 - x/2
= [cos(x/2) - sin(x/2)] [cos(x/2) + sin(x/2)] / [sin(x/2)+cos(x/2)]2
= [cos(x/2) - sin(x/2)] / [cos(x/2) + sin (x/2)]
= 1 - tan(x/2) / 1+tan(x/2)
= tan (pi/4 - x/2)
=tan-1[tan(pi/4 - x/2)]
= pi/4 - x/2
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