Chứng minh đẳng thức sau(giả sử biểu thức sau có nghĩa)

Có `\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4sin^2x}`

`=\sqrt{(sin^2x+4(1-sin^2x)}+\sqrt{cos^4x+4(1-cos^2x)}`

`=\sqrt{sin^2x-4sin^2x+4}+\sqrt{cos^4x-4cos^2x+4}`

`=\sqrt{(sin^2x-2)^2}+\sqrt{(cos^2x-2)^2}`

`=2-sin^2x+2-cos^2x` (vì `sin^2x; cos^2x<=1`)

`=4-(sin^2x+cos^2x)`

`=4-1`

`=3`

Có `3tan(x+(\pi)/(3))tan((\pi)/(6)-x)`

`=3(sin(x+(\pi)/(3)))/(cos(x+(\pi)/(3)))(sin((\pi)/(6)-x))/(cos((\pi)/(6)-x))`

Có `sin(x+(\pi)/(3))sin((\pi)/(6)-x)`

`=(1)/(2)[cos(x+(\pi)/(3)-(\pi)/(6)+x)-cos(x+(\pi)/(3)+(\pi)/(6)-x)]`

`=(1)/(2)[cos(2x+(\pi)/(6))-cos  (\pi)/(2)]`

`=(1)/(2)cos(2x+(\pi)/(6))`

Tương tự ta có `cos(x+(\pi)/(3))cos((\pi)/(6)-x)=(1)/(2)cos(2x+(\pi)/(6))`

`-> 3tan(x+(\pi)/(3))tan((\pi)/(6)-x)=3.1=3`

`-> \sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4sin^2x}=3tan(x+(\pi)/(3))tan((\pi)/(6)-x) (đpcm)`