Help with trig identities?
Im stuck on this one..
(secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA)
- Snugglebunnie101Lv 52 years agoFavorite Answer
OK, here you go sweetheart :)
Note the identities used in the proof :
secA = 1/cosA
tanA = sinA/cosA
1 - sin²A = cos²A
Starting with the LHS we have :
= [(secA + tanA) /(secA - tanA)] - [(1 - sinA) /(1 + sinA)
= [(1/cosA + sinA/cosA) /(1/cosA - sinA/cosA)] - [(1 - sinA) /(1 + sinA)]
= [[(1 + sinA) /cosA] ÷ [(1 - sinA) /cosA]] - [(1 - sinA) /(1 + sinA)]
= [(1 + sinA) /(1 - sinA)] - [(1 - sinA) /(1 + sinA)]
= [(1 + sinA)² - (1 - sinA)²] /(1 - sin²A)
= [1 + 2sinA + sin²A - 1 + 2sinA - sin²A] /(1 - sin²A)
= [4sinA] /cos²A
= 4 (sinA/cosA) (1/cosA)
= RHS, as required, Q.E.D.
Oh, I forgot to mention that what you've written in your question is NOT an identity! It should be a 'secA' instead of a 'cosecA' at the end for the identity to be valid.
Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
P.S. (Don't forget to vote me best answer as being the first to correctly and thoroughly answer your question!)
- PopeLv 72 years ago
Take it or leave it, but on these identity proofs I always begin by checking whether there are any conflicts in the domains of the two sides of the equation. In this case there are.
Let A = 0.
The left side of the equation is zero. The right side is undefined. When a real number is equated with an undefined expression, the equation cannot be an identity. And no, there is no such thing as expressions that are almost an identical.
- TomVLv 72 years ago
The reason you are stuck on that one may be because it is not an identity.
Let A = 60°
sinA = √3/2
secA = 1/cosA = 2
cscA = 1/sinA = 2/√3
tanA = √3
(secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA) = 4tanAcscA
(2 + √3)/(2 - √3) - (1-√3/2)/(1+√3/2) = 4(√3)(2/√3)
(2+√3)/(2-√3) - (2-√3)/(2+√3) = 8
(2+√3)²/(4-3) - (2-√3)²/(4-3) = 8
4 + 4√3 + 3 - (4 - 4√3 + 3) = 8
7 + 4√3 - 7 + 4√3 = 8
8√3 = 8 : False statement
Equation is not an identity.
(secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA) = 4tanAsecA
is an identity. (Note the different RHS)
LHS = (secA+tanA)/(secA-tanA) - (1-sinA)/(1+sinA)
replace secA with 1/cosA and tanA with sinA/cosA
= (1/cosA + sinA/cosA)/(1/cosA - sinA/cosA) - (1-sinA)/(1+sinA)
multiply numerator and denominator of first term by cosA
= (1+sinA)/(1-sinA) - (1-sinA)/(1+sinA)
multiply numerator and denominator of first term by (1+sinA) and second term by (1-sinA)
= (1+sinA)²/(1-sin²A) - (1-sinA)²/(1 - sin²A)
replace 1 - sin²A with cos²A per Pythagorean Identity, and combine terms with common denominator
= [(1+sinA)² - (1-sinA)²]/cos²A
expand squared terms
= [(1 + 2sinA + sin²A - (1 - 2sinA + sin²A)]/cos²A
combine like terms
replace sinA/cosA with tanA and 1/cosA with secA
= RHS (as modified)
- 2 years ago
Try multiplying the fractions by their conjugate (sec A + tan A) or (sec A - tan A) and (1 - sin A) or (1 + sin A) to create identities (sec²A - tan²A = 1) and (1 - sin²A = cos²A)