(sin^2A-cos^2A)(1-sinAcosA)/cosA(secA-cosA)(sin^3A+cos^3A)=sinA
thank you for helping
2007-04-30 01:36:43 · 3 answers · asked by jin_ale 2 in Science & Mathematics ➔ Mathematics
(sin^2A-cos^2A)(1-sinAcosA)/cosA(secA-cosA)(sin^3A+cos^3A)=sinA
We simplify only the left hand side.
=> (sin^2A-cos^2A)(sin^2A-sinAcosA+cos^2A)/cosA(secA-cosA)(sin^3A+cos^3A) [since sin^2A + cos^2A = 1]
=> (sin^A-cos^A)(sin^A+cos^A)(sin^2A-sinAcosA+cos^2A)/cosA(secA-cosA)(sin^3A+cos^3A)
=> (sin^A-cos^A)(sin^3A+cos^3A)/cosA(secA-cosA)(sin^3A+cos^3A) [since (a+b)(a^2-ab+b^2) = a^3+b^3]
=> (sin^A-cos^A)/cosA(secA-cosA)
=> (sin^A-cos^A)/(1 - cos^2A) [since cosAsecA = 1]
=> (sin^A-cos^A)/sin^2A [since sin^2A + cos^2A = 1]
...well we see this one wont simplify to sinA, maybe there is error on my solution or the original identity is incomplete. Still, some of the methods here might be useful to your problem at hand.
2007-04-30 02:08:09 · answer #1 · answered by wala_lang 2 · 0⤊ 0⤋
(sin^2A-cos^2A)(1-sinAcosA)/co...
Nobody understands this. What is the denominator co... ?
2007-04-30 09:00:10 · answer #2 · answered by Robert L 7 · 0⤊ 0⤋
denominator co?? incomplete
2007-04-30 09:13:13 · answer #3 · answered by mustafa k 2 · 0⤊ 0⤋