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Hero Group TitleBest ResponseYou've already chosen the best response.1
\[\space \space \space \sec^2x \div (\cot^2x + 1)\] \[=\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + 1\right)\] \[=\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + \frac{\sin^2x}{\sin^2x}\right)\] \[=\frac{1}{\cos^2x} \div \left(\frac{1}{\sin^2x} \right)\] \[=\frac{1}{\cos^2x} \times \left(\frac{\sin^2x}{1} \right)\] \[=\frac{\sin^2x}{\cos^2x}\] \[=\tan^2x\]
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
in my txbook the answer is tan^2x
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
There's a negative in front that cannot be ignored
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
Are you sure you wrote down the problem correctly?
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
k it was sec not sec sorry i just wrote it : instead of writing it :
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
can u plz help me with other one?
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
cotxcos^3xcscx
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
\[\cot x  \cos^3x \csc x \\=\frac{\cos x}{\sin x}  \frac{\cos^3x}{\sin x} \\=\frac{\cos x  \cos^3x}{\sin x} \\=\frac{\cos x(1  \cos^2x)}{\sin x} \\=\frac{\cos x(\sin^2x)}{\sin x} \\=\cos x \sin x \]
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
thank you so much u rock!i wish u were my pre cal teacher
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
wait how did u get cos^3xcosx=cosx(1cos^2x) ?
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
can't we just subtract it out?
 one year ago

drasti1211 Group TitleBest ResponseYou've already chosen the best response.0
k i get what u did !thanks for all ur help
 one year ago
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