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```                                                   2sinß

1 + sinß + cosß

= k

then find
1 + sinß - cosß

1+ sinß
```
8 years ago

6 Points
```										Given : 2 sinß / (1 + sinß + cosß) = k
Multiply numerator & denominator by (1 + sinß - cosß)
We get : 2 sinß (1 + sinß - cosß) /  (1 + sinß + cosß) (1 + sinß - cosß)
Denominator is of the form (a + b) (a - b) which is (a^2 - b^2)
Hence this further becomes : 2 sinß (1 + sinß - cosß) / ((1 + sinß)^2 - cos^2ß)
Expanding the denominator, we get : 2 sinß (1 + sinß - cosß) / (1 + sin^2ß + 2sinß - cos^2ß)
Substitute 1 with sin^2ß + cos^2ß in the denominator : 2 sinß (1 + sinß - cosß) / (sin^2ß + cos^2ß + sin^2ß + 2sinß - cos^2ß)
Cancelling cos^2ß - cos^2ß, we get : 2 sinß (1 + sinß - cosß) / (sin^2ß + sin^2ß + 2sinß)
Further simplify as : 2 sinß (1 + sinß - cosß) / (2sin^2ß +  2sinß)
=> 2 sinß (1 + sinß - cosß) / 2 sinß (1 +  sinß)
Cancelling 2 sinß from both numerator & denominator, we get : (1 + sinß - cosß) / (1 +  sinß)
Since the given equation is simplified to this form, the value of (1 + sinß - cosß) / (1 +  sinß) IS ALSO "k"
Solution : Given  2 sinß / (1 + sinß + cosß) = k, then  (1 + sinß - cosß) / (1 +  sinß) = k

```
8 years ago
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