Proving equations (trig)

[From: ] [author: ] [Date: 11-10-02] [Hit: ]

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1 - cos^2A/1 + sinA = sinA
multiply both sides by 1 + sinA: 1 + sinA - cos^2A = sinA*(1+sinA)
multiply through: 1 + sinA - cos^2A = sinA + sin^2A
subtract sinA from both sides: 1 - cos^2A = sin^2A
add cos^2A to both sides: 1 = sin^2A + cos^2A, this is the same trig identity we obtained before.

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Is it okay if I remove the angles?

tan^2A - sin^2A = (sin^2/cos^2) - ((cos^2*sin^2)/cos^2) = (sin^2 - sin^2*cos^2)/cos^2 = ((1-cos^2)*sin^2)/cos^2) = sin^2*sin^2/cos^2 = tan^2*sin^2

sinA * cosA * (tanA + cotA) = sinA(cosA*tanA + cosA*cotA) = sin(cos*(sin/cos) + cos*(cos/sin)) = sin*(sin + cos^2/sin) = sin^2 + cos^2 = 1

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tan^2A - sin^2A
= sin^2A[sec^2A -- 1]
= sin^2A * tan^2A
= tan^2A * sin^2A

sinA/cosA * (1 - cot^2A) + cosA/sinA * (1 - tan^2A)
= sin^2A*(1 -- cot^2A) + cos^2A*(1 -- tan^2A)
= sin^2A -- cos^2A + cos^2A -- sin^2A
= 0

sinA * cosA * (tanA + cotA)
= sinA*cosA*(sinA/cosA + cosA/sinA)
= sin^2A + cos^2A
= 1

1 - cos^2A/1 + sinA = sinA THIS IS WRONG

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keywords: trig,equations,Proving,Proving equations (trig)