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How would I prove this trig identity: (1+cos^2x-2cosx)/sin^2x=(1-cosx)/(1+cosx)
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See attached file for a clear version
hmm what "solve" stand for?
Sorry I meant "prove"
wel.. let's take a peek at the left-hand-side notice the numerator.. can you factor it?
That's what I'm stuck on, I don't get how I can factor the numerator.
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ok... let us take a closer look at it \(\bf \cfrac{1+cos^2(x)-2cos(x)}{sin^2(x)}=\cfrac{1-cos(x)}{1+cos(x)} \\ \quad \\ 1+cos^2(x)-2cos(x)\implies cos^2(x)-2cos(x)+1 \\ \quad \\\ [{\color{brown}{ cos(x)}}]^2-2{\color{brown}{ cos(x)}}+1\implies (\square ?-\square ?)(\square ?-\square ?)\)
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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