SOLUTION: Hi. Please help me verify this equation as an identity: {{{sec(theta)/tan(theta)}}} + {{{(1 - tan(theta))/sec(theta)}}} = {{{(1+tan(theta))/(sec(theta)tan(theta))}}} I have

Algebra ->  Trigonometry-basics -> SOLUTION: Hi. Please help me verify this equation as an identity: {{{sec(theta)/tan(theta)}}} + {{{(1 - tan(theta))/sec(theta)}}} = {{{(1+tan(theta))/(sec(theta)tan(theta))}}} I have      Log On


   



Question 602442: Hi. Please help me verify this equation as an identity:
sec%28theta%29%2Ftan%28theta%29 + %281+-+tan%28theta%29%29%2Fsec%28theta%29 = %281%2Btan%28theta%29%29%2F%28sec%28theta%29tan%28theta%29%29

I have been trying to work on this for an hour now but I always end up with 1/sin theta = 1, or csc theta = 1, and obviously that's not the right answer.
Please show me how to do this properly since my teacher only speaks to the white board...
Thank you!

Found 2 solutions by Alan3354, AnlytcPhil:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sec theta/tan theta + (1 - tan theta)/sec theta = (1+tan theta)/(sec theta)(tan theta)
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sec/tan + (1 - tan)/sec = (1 + tan)/(sec*tan) (I'll assume the last den is sec*tan)
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Multiply thru by sec
sec^2/tan + 1 - tan = (1 + tan)/tan = cot + 1
sec^2/tan + 1 - tan = cot + 1
Multiply by tan
sec^2 + tan = 1 + tan
Not an identity.
====================
You need to make it clear, add parentheses to eliminate guessing at what the terms are.

Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
sec%28theta%29%2Ftan%28theta%29 + %281+-+tan%28theta%29%29%2Fsec%28theta%29 = %281%2Btan%28theta%29%29%2F%28sec%28theta%29tan%28theta%29%29

The LCD on the left side is sec(θ)tan(θ)

expr%28sec%28theta%29%2Ftan%28theta%29%29%2Aexpr%28sec%28theta%29%2Fsec%28theta%29%29 +  =

expr%28sec%5E2%28theta%29%2F%28tan%28theta%29sec%28theta%29%29%29 + expr%28%28tan%28theta%29+-+tan%5E2%28theta%29%29%2Fsec%28theta%29tan%28theta%29%29 =

 =

[Next use the identity 1 + tanē(θ) = secē(θ) which is equivalent
to secē(θ) - tanē(θ) = 1, and you have:]

%281%2Btan%28theta%29%29%2F%28sec%28theta%29tan%28theta%29%29

Edwin