Question

How derivative of (1/cot(x)) = sec^2(x) ?

Answer 1

`d/dx 1/cotx = d/dxtgx = sec^2x` because, `tgx = sinx/cosx`, like this you have to derive using the rule of the quotient. See bellow:

Explanation:

`d/dx tgx = d/dx sinx/cosx = (sinx' cosx - sinx cosx')/(cosx)^2 = (cosxcosx - sinx(-sinx))/(cosx)^2 = (cos^2x + sin^2x)/(cosx)^2 = 1/ cos^2x = sec^2x`.

I hope to helped you!

Answer 2

the answer
`tan^2x+1=2sec^2x*tanx`

Explanation:

show below

`d/dx[1/cotx]=d/dx[sec^2x]`

`cotx=cosx/sinx`

`d/dx[sinx/cosx]=d/dx[sec^2x]`

`[cosx*(cosx)-sinx*(-sinx)]/cos^2x=2secx*secx*tanx`

`[cos^2x+sin^2x]/cos^2x=2sec^2x*tanx`

`(sin^2x/cos^2x)+(cos^2x/cos^2x)=2sec^2x*tanx`

`tan^2x+1=2sec^2x*tanx`