# Prove secA-cosA=tanAsinA 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:

`secA-cosA=1/cosA-cosA`

`` 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:

`1-cos^2A=sin^2A`

Finally, `(sin^2A)/cosA=sinA/cosA*sinA=tanAsinA`

`` which is equivalent to the right hand side of the identity you had to prove.

secA = 1/cosA

i.e..,

secA-cosA=[1/cosA]-cosA

=[1-cosA*cosA]/cosA

=[1-cos^2A]/cosA (since 1-cos^2A= SinA)

=sin^2A/cosA

=sinA*sinA/cosA

=[sinA/cosA]*sinA

=TanA*sinA