Question

Question #b314f

Answer 1

`1/(sinxcosx)-tanx=cotx`

Explanation:

`1/(sinxcosx)-tanx`

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

`=1/(sinxcosx)-sinx/cosxcolor(red)(*sinx/sinx)`

`=frac{1-sin^2x}{sinxcosx}`

(Pythagorean identity in the numerator)
`=frac{cos^2x}{sinxcosx}`

`=frac{cosx}{sinx}`

`=cotx`