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# 2sinß                                    1 + sinß + cosß= kthen find                            1 + sinß - cosß                                 1+ sinß

12 years ago

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^2ßcos^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