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

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

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

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

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

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

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

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

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

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