Question

Question #849cc

Answer 1

`1/(sinxcosx)-sinx/cosx=cotx`

Explanation:

`1/(sinxcosx)-sinx/cosx`=`1/cosx(1/sinx-sinx)`

=`1/cosx(1/sinx-sin^2x/sinx)`
=`1/(cosxsinx)(1-sin^2x)`

But `1-sin^2x=cos^2x`

So the expression can be written as
`cos^2x/(cosxsinx)`

Cancelling gives
`cosx/sinx`=`cotx`