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drasti1211
simplify:-sec^2x/cot^2x+1
\[\space \space \space -\sec^2x \div (\cot^2x + 1)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + 1\right)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + \frac{\sin^2x}{\sin^2x}\right)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{1}{\sin^2x} \right)\] \[=-\frac{1}{\cos^2x} \times \left(\frac{\sin^2x}{1} \right)\] \[=-\frac{\sin^2x}{\cos^2x}\] \[=-\tan^2x\]
in my txbook the answer is tan^2x
There's a negative in front that cannot be ignored
Are you sure you wrote down the problem correctly?
k it was sec not -sec sorry i just wrote it :- instead of writing it :
can u plz help me with other one?
\[\cot x - \cos^3x \csc x \\=\frac{\cos x}{\sin x} - \frac{\cos^3x}{\sin x} \\=\frac{\cos x - \cos^3x}{\sin x} \\=\frac{\cos x(1 - \cos^2x)}{\sin x} \\=\frac{\cos x(\sin^2x)}{\sin x} \\=\cos x \sin x \]
thank you so much u rock!i wish u were my pre cal teacher
wait how did u get cos^3x-cosx=cosx(1-cos^2x) ?
can't we just subtract it out?
k i get what u did !thanks for all ur help