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

```
5 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)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 . :)
```
5 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 – Y2   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)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
```
5 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)
```
5 years ago
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