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cosA / 1 - sinA = tan (45 + A/2)

cosA / 1 - sinA = tan (45 + A/2)

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3 Answers

Sam Abraham
30 Points
6 years ago
CosA / 1 – SinA can be written as  
Cos2A/2 – Sin2A/2 / Cos2A + Sin2A/2 +2Sin A/2 Cos A/2 ….…
which is equal to = Cos2A/2 – Sin2A/2 / (CosA/2 + SinA/2)2 .. As Sin2A/2 +2Sin A/2 Cos A/2 is in the form of X2 + Y2 +2XY = (X+Y)2  
And Cos2A/2 – Sin2A can be written as  (CosA/2 – SinA/2) (CosA/2 + SinA/2) .. as it is in the form of (X + Y)(X – Y) = X2 – Y2  
So Cos2A/2 – Sin2A/2 / Cos2A + Sin2A/2 +2Sin A/2 Cos A/2 =
(CosA/2 – SinA/2) (CosA/2 + SinA/2) / (CosA/2 + SinA/2)2
(CosA/2 – SinA/2) (CosA/2 + SinA/2) / (CosA/2 + SinA/2) = 
(CosA/2 – SinA/2) / (CosA/2 + SinA/2) .
Now divide the numerator and the denominator by CosA/2 .. because of which we get ..
1 – tan A/2 / 1 + tan A/2 = tan 45 + tan A/2 / 1 – tan 45 * tan A/2 = tan (45 +A/2)  .
Guess it helped . :) 
 
 
 
Sam Abraham
30 Points
6 years ago
 
CosA / 1 – SinA can be written as  
Cos2A/2 – Sin2A/2 / Cos2A/2 + Sin2A/2 +2Sin A/2 Cos A/2 ….…
which is equal to = Cos2A/2 – Sin2A/2 / (CosA/2 + SinA/2)2 .. As Sin2A/2 + Cos2A/2 + 2Sin A/2 Cos A/2 is in the form of X2 + Y2 +2XY = (X+Y)2 
And Cos2A/2 – Sin2A/2 can be written as  (CosA/2 – SinA/2) (CosA/2 + SinA/2) .. as it is in the form of (X + Y)(X – Y) = X2 – Y  
So Cos2A/2 – Sin2A/2 / Cos2A/2 + Sin2A/2 +2Sin A/2 Cos A/2 =
(CosA/2 – SinA/2) (CosA/2 + SinA/2) / (CosA/2 + SinA/2)2 = 
(CosA/2 – SinA/2) (CosA/2 + SinA/2) / (CosA/2 + SinA/2) = 
(CosA/2 – SinA/2) / (CosA/2 + SinA/2) .
Now divide the numerator and the denominator by CosA/2 .. because of which we get ..
1 – tan A/2 / 1 + tan A/2 = tan 45 + tan A/2 / 1 – tan 45 * tan A/2 = tan (45 +A/2)  .
Guess it helped . :) 
Revised Answer
grenade
2061 Points
6 years ago
 Use the RHS as a hint how to proceed. 

sinA = 2sin(A/2)cos(A/2) 
cosA = 2cos^2 (A/2) - 1 

sinA / (1+cosA) = sin(A/2) / cos(A/2) = tan(A/2)

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