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OpenStudy (anonymous):
is sin x + 1 equivalent to sin2x−sinx−2/sinx−2?
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OpenStudy (anonymous):
\[is~\it~\sin^2x~or ~\sin2x\]
OpenStudy (amoodarya):
\[\frac{\sin^2x-sinx-2}{sinx-2}=\frac{(sinx+1)(sinx-2)}{sinx-2}\\sinx-2\neq0\rightarrow \\\frac{(sinx+1)(sinx-2)}{sinx-2}=sinx +1\]
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