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OpenStudy (anonymous):
Simplify the expression. sin^2x-1/cos(-x)
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OpenStudy (anonymous):
@hartnn
hartnn (hartnn):
is that \(\Large \dfrac{\sin^2x-1}{\cos (-x)}\)
OpenStudy (anonymous):
yes
hartnn (hartnn):
do you know \(\sin^2x + \cos^2x = 1\) ?
OpenStudy (anonymous):
thank you both
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