Question

simplify:-sec^2x/cot^2x+1

Answer 1

\[\space \space \space -\sec^2x \div (\cot^2x + 1)\]
\[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + 1\right)\]
\[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + \frac{\sin^2x}{\sin^2x}\right)\]
\[=-\frac{1}{\cos^2x} \div \left(\frac{1}{\sin^2x} \right)\]
\[=-\frac{1}{\cos^2x} \times \left(\frac{\sin^2x}{1} \right)\]
\[=-\frac{\sin^2x}{\cos^2x}\]
\[=-\tan^2x\]

Answer 2

in my txbook the answer is tan^2x

Answer 3

There's a negative in front that cannot be ignored

Answer 4

Are you sure you wrote down the problem correctly?

Answer 5

k wait

Answer 6

k it was sec not -sec sorry i just wrote it :- instead of writing it :

Answer 7

can u plz help me with other one?

Answer 8

cotx-cos^3xcscx

Answer 9

\[\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

\]

Answer 10

thank you so much u rock!i wish u were my pre cal teacher

Answer 11

okay

Answer 12

wait how did u get cos^3x-cosx=cosx(1-cos^2x) ?

Answer 13

can't we just subtract it out?

Answer 14

k i get what u did !thanks for all ur help