## anonymous 3 years ago simplify:-sec^2x/cot^2x+1

1. Hero

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

2. anonymous

in my txbook the answer is tan^2x

3. Hero

There's a negative in front that cannot be ignored

4. Hero

Are you sure you wrote down the problem correctly?

5. anonymous

k wait

6. anonymous

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

7. anonymous

can u plz help me with other one?

8. anonymous

cotx-cos^3xcscx

9. Hero

$\cot x - \cos^3x \csc x \\=\frac{\cos x}{\sin x} - \frac{\cos^3x}{\sin x} \\=\frac{\cos x - \cos^3x}{\sin x} \\=\frac{\cos x(1 - \cos^2x)}{\sin x} \\=\frac{\cos x(\sin^2x)}{\sin x} \\=\cos x \sin x$

10. anonymous

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

11. Hero

okay

12. anonymous

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

13. anonymous

can't we just subtract it out?

14. anonymous

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