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

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

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

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

drasti1211
 2 years ago
Best ResponseYou've already chosen the best response.0k it was sec not sec sorry i just wrote it : instead of writing it :

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

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

drasti1211
 2 years ago
Best ResponseYou've already chosen the best response.0thank you so much u rock!i wish u were my pre cal teacher

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

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

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