Question

Question #3c29b

Answer 1

is it 1-tan^2x instead of 1-tanx?

Explanation:

`1/cos^2x = 1/(1-sin^2x)`
`1/(1-sin^2x) = 1-1/sin^2x`
`1-1/(1-cos^2x) = 1-1-sec^2x`
`1-1-sec^2x = sec^2x = 1-tan^2x`

Answer 2

`sinx=0` or `sinx+cosx=0`

`x=0, (3pi)/4, pi, (7pi)/4`

Explanation:

`1/cos^2x=1-tanx`

`1=cos^2x(1-tanx)`

`1=cos^2x-tanxcos^2x`

`1=cos^2x-sinx/cosxcos^2x`

`1=cos^2x-sinxcosx`

`sin^2x+cos^2x=cos^2x-sinxcosx`

`sin^2x=-sinxcosx`

`sin^2x+sinxcosx=0`

`sinx(sinx+cosx)=0`

`sinx=0` or `sinx+cosx=0`
`sinx=0` or `sinx=-cosx`

Edit: from the comments:

For the range `0<=x<=2pi`

a) `sinx=0`
`:.x=0, x=pi`

b) `sinx+cosx=0`
We are going to divide both sides by `cosx`. It is necessary to check beforehand that we are not dividing by zero. However, if `cosx=0, sinx=1`, and the sum is not zero. In this case, we are safe to divide by `cosx`/

`sinx/cosx+cosx/cosx=0`
`tanx+1=0`
`tanx=-1`

`x=(3pi)/4, (7pi)/4`

Putting this together:

`x=0, (3pi)/4, pi, (7pi)/4`

Answer 3

`tanx=0 and tanx=-1`

Explanation:

`1/cos^2x=1-tanx`

Identities:

`color(red)(1/cos^2x=sec^2x)`

`color(red)(sec^2x=1+tan^2x)`

So we have:

`1+tan^2x=1-tanx`

Subtract `1` from both sides:

`tan^2x=-tanx`

`tan^2x+tanx=0`

Factor:

`tanx(tanx+1)=0`

`tanx=0 and tanx=-1`

You haven't given an interval, so for `0<=x<=2pi`

`0, pi , (3pi)/4,(7pi)/4`

Graph: