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Mathematics 23 Online
OpenStudy (anonymous):

How would I prove this trig identity: (1+cos^2x-2cosx)/sin^2x=(1-cosx)/(1+cosx)

OpenStudy (anonymous):

See attached file for a clear version

OpenStudy (jdoe0001):

hmm what "solve" stand for?

OpenStudy (anonymous):

Sorry I meant "prove"

OpenStudy (jdoe0001):

wel.. let's take a peek at the left-hand-side notice the numerator.. can you factor it?

OpenStudy (anonymous):

That's what I'm stuck on, I don't get how I can factor the numerator.

OpenStudy (jdoe0001):

ok... let us take a closer look at it \(\bf \cfrac{1+cos^2(x)-2cos(x)}{sin^2(x)}=\cfrac{1-cos(x)}{1+cos(x)} \\ \quad \\ 1+cos^2(x)-2cos(x)\implies cos^2(x)-2cos(x)+1 \\ \quad \\\ [{\color{brown}{ cos(x)}}]^2-2{\color{brown}{ cos(x)}}+1\implies (\square ?-\square ?)(\square ?-\square ?)\)

OpenStudy (jdoe0001):

notice, the numerator of the left-side is just a quadratic equation, thus you could just factor it like any quadratic one

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