We have to prove that (sin A + cos B)/(sin A - cos B) = (sec B + csc A)/(sec B - csc A)

We use the definitions: sec x = 1/cos x and csc x = 1/ sin x

(sin A + cos B)/(sin A - cos B)

=>...

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We have to prove that (sin A + cos B)/(sin A - cos B) = (sec B + csc A)/(sec B - csc A)

We use the definitions: sec x = 1/cos x and csc x = 1/ sin x

(sin A + cos B)/(sin A - cos B)

=> (1/ csc A + 1/sec B)/ (1/csc A - 1/sec B)

making the denominator the same

=> [(sec B + csc A)/(csc A * sec B)]/[(sec B - csc A)/(csc A * sec B)]

canceling the common denominator

=> [(sec B + csc A)]/[(sec B - csc A)]

**This proves that (sin A + cos B)/(sin A - cos B) = (sec B + csc A)/(sec B - csc A)**