Must show all work. HELP PLEASE?
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Let us try to prove that left side is equal to the right side.
We know that secA = 1/cosA.
So, we will have:
`1/(cosA) - cosA`
Hence, lcd = cosA. We multiply top and bottom of cosA by cosA.
`1/(cosA) - (cosA*cosA)/cosA = 1/(cosA) - (cos^2A)/cosA = (1 - (cos^2A))/cosA`
We know that sin^2A = 1 - cos^2A.
`(1 - (cos^2A))/(cosA) = (sin^2A)/cosA`
We can rewrite that:
`(sin^2A)/(cosA) = (sinA/cosA)*sinA`
Take note that tanA = sinA/cosA.
`(sinA/cosA)*sinA = tanAsinA`
When working with trig identities, rewriting secant as a reciprocal of a cosine is a good place to start:
`` Then, perform the subtraction by rewriting cosA as a fraction with the denominator cosA:
`1/cosA - cosA = 1/cosA - (cos^2A)/cosA=(1-cos^2A)/cosA`
`` Now use the Pythagorean identity to rewrite the numerator:
`` which is equivalent to the right hand side of the identity you had to prove.
secA = 1/cosA
=[1-cos^2A]/cosA (since 1-cos^2A= SinA)
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