# Prove that: sinA+cosB/sinA-cosB=secB+cscA/secB-cscA

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### 2 Answers

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)**

Supposing that we have to prove that (sinA+cosB)/(sinA-cosB) = (secB+cscA)/(secB-cscA), we'll cross multiply and we'll get:

(sinA+cosB)(secB-cscA) = (sinA-cosB)(secB+cscA)

We'll pu sec B = 1/cos B and sec B = 1/cos B

csc A = 1/sin A and sec A = 1/cos A

We'll substitute secA,secB,cscA and cscB inside brackets:

(sinA+cosB)(1/cos B - 1/sin A) = (sinA-cosB)(1/cos B + 1/sin A)

We'll remove brackets using FOIL method:

sinA/cosB - sinA/sinA + cosB/cosB - cosB/sinA = sinA/cosB + sinA/sinA - cosB/cosB - cosB/sinA

sinA/cosB -1 + 1- cosB/sinA = sinA/cosB + 1- 1 - cosB/sinA

We'll eliminate like terms and we'll get LHS = RHS:

**sinA/cosB - cosB/sinA = sinA/cosB - cosB/sinA**