Question

Question #4352a

Answer 1

`1`

Explanation:

If `3sin^2x-sin^4x=1` then `sin^2x=(3-sqrt(5))/2`

and

`cos^2x=1-sin^2x=1-(3-sqrt(5))/2`

then

`tan^2x+tan^4x =((3-sqrt(5))/2)/(1-(3-sqrt(5))/2)+(((3-sqrt(5))/2)/(1-(3-sqrt(5))/2))^2=1`

Answer 2

`3 - sin^2 x`

Explanation:

Develop the left side of the equation:
`3sin^2 x - sin^4 x = 1`
`sin^2 x(3 - sin^2 x) = 1`
From this equation we get:
`1/sin^2 x = 3 - sin^2 x` (1)

Develop the right side:
`tan^2 x + tan^4 x = tan^2 x(1 + tan^2 x) = tan^2 x(1/cos^2 x) =`
`= (cos^2 x)/(sin^2 x)(1/(cos^2 x)) = 1/(sin^2 x) (2)`
Compare (1) and (2) -->
`tan^2 x + tan^4 x = 3 - sin^2 x`

Answer 3

`3sin^2x-sin^4x=1`

Dividing bothsides by `cos^4x`

`=>(3sin^2x)/cos^4x-sin^4x/cos^4x=1/cos^4x`

`=>3tan^2xsec^2x-tan^4x=sec^4x`

`=>3tan^2x(1+tan^2x)-tan^4x=(1+tan^2x)^2`

`=>3tan^2x+3tan^4x-tan^4x=1+2tan^2x+tan^4x`

`=>3tan^2x-2tan^2x+3tan^4x-tan^4x-tan^4x=1`

`=>tan^2x+tan^4x=1`