Yahoo Answers: Answers and Comments for Help with trig identities? [Mathematics]
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From Anonymous
enUS
Sun, 19 Nov 2017 05:59:23 +0000
3
Yahoo Answers: Answers and Comments for Help with trig identities? [Mathematics]
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From Snugglebunnie101: OK, here you go sweetheart :)
Note the ident...
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Sun, 19 Nov 2017 06:21:24 +0000
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 :
∴ LHS
= [(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)
= 4(tanA)(secA)
= 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!)

From Pope: Take it or leave it, but on these identity pro...
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Sun, 19 Nov 2017 07:46:38 +0000
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.

From TomV: The reason you are stuck on that one may be be...
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Sun, 19 Nov 2017 06:47:32 +0000
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)/(secAtanA)  (1sinA)/(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/(43)  (2√3)²/(43) = 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.
However
(secA+tanA)/(secAtanA)  (1sinA)/(1+sinA) = 4tanAsecA
is an identity. (Note the different RHS)
LHS = (secA+tanA)/(secAtanA)  (1sinA)/(1+sinA)
replace secA with 1/cosA and tanA with sinA/cosA
= (1/cosA + sinA/cosA)/(1/cosA  sinA/cosA)  (1sinA)/(1+sinA)
multiply numerator and denominator of first term by cosA
= (1+sinA)/(1sinA)  (1sinA)/(1+sinA)
multiply numerator and denominator of first term by (1+sinA) and second term by (1sinA)
= (1+sinA)²/(1sin²A)  (1sinA)²/(1  sin²A)
replace 1  sin²A with cos²A per Pythagorean Identity, and combine terms with common denominator
= [(1+sinA)²  (1sinA)²]/cos²A
expand squared terms
= [(1 + 2sinA + sin²A  (1  2sinA + sin²A)]/cos²A
combine like terms
= 4sinA/cos²A
replace sinA/cosA with tanA and 1/cosA with secA
= 4tanAsecA
= RHS (as modified)

From Physnitch: Try multiplying the fractions by their conjuga...
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Sun, 19 Nov 2017 06:17:23 +0000
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)