Question

Im stuck on trying to prove: 2cos^2x-sin^2x+1=3cos^2x & tanx(sinx+cotx*cosx)=secx can anyone help??

Answer 1

See below

Explanation:

First one using: `sin^2x=1-cos^2x`
Start :
`2cos^2x-sin^2x+1=3cos^2x`

`2cos^2x-(1-cos^2x)+1=3cos^2x`

`2cos^2xcancel(-1)+cos^2xcancel(+1)= 3cos^2x`

`2cos^2x+cos^2x= 3cos^2x`

`3cos^2x= 3cos^2x`

End

Second one using:
`cotx=cosx/sinx`
`tanx=sinx/cosx`
`sin^2x+cos^2x=1`
`1/cosx=secx`

Start :
`tanx(sinx+cotx*cosx)=secx`

`tanx(sinx+cosx/sinx*cosx)= secx`

`sinx/cosx(sinx+cos^2x/sinx)= secx`

`sinx/cosx*sinx+sinx/cosx*cos^2x/sinx= secx`

`sin^2x/cosx+cosx= secx`

`sin^2x/cosx+cos^2x/cosx= secx`

`(sin^2x+cos^2x)/cosx= secx`

`1/cosx= secx`

`secx=secx`
End