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Simplify (sin^2x+sinx-6)/sinx+3
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\[\frac{ \sin^2x+sinx-6 }{sinx+3}\]
Let \(\bf sin(x) = u\), then:\[\bf \frac{ u^2+u-6 }{ u+3 }=\frac{\cancel{(u+3)}(u-2)}{\cancel{u+3}}=u-2\]At this point, we can replace u with sin(x) again and we get:\[\bf =u-2=\sin(x)-2\]And we are done with the simplification. @Jay567
Than you! That's exactly what my book has. Trig proofs are tricky for me!
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