Question

Sec40+ sec20 tan2 0 - 2 tan4 0 =3 sec² 0 -2Sect0-210

Answer 1

Given:

`sec^4\theta+sec^2\theta tan^2\theta-2tan^4\theta=3sec^2\theta-2`

Required:

We need to prove the given equation.

Step-by-step explanation:

Consider the left-hand side of the equation.

`Add\text{ and subtract }3tan^4\theta.`

`sec^4\theta+sec^2\theta tan^2\theta-2tan^4\theta=sec^4\theta+sec^2\theta tan^2\theta-2tan^4\theta+3tan^4\theta-3tan^4\theta`
`=sec^4\theta+sec^2\theta tan^2\theta+tan^4\theta-3tan^4\theta`

`Add\text{ and subtract -2}sec^2\theta tan^2\theta.`

`=sec^4\theta+sec^2\theta tan^2\theta+tan^4\theta-3tan^4\theta-2sec^2\theta tan^2\theta+2sec^2\theta tan^2\theta`
`=sec^4\theta-2sec^2\theta tan^2\theta+tan^4\theta-3tan^4\theta+sec^2\theta tan^2\theta+2sec^2\theta tan^2\theta`
`=sec^4\theta-2sec^2\theta tan^2\theta+tan^4\theta-3tan^4\theta+3sec^2\theta tan^2\theta`
`Use\text{ }sec^4\theta-2sec^2\theta tan^2\theta+tan^4\theta=(sec^2\theta-tan^2\theta)^2`
`=(sec^2\theta-tan^2\theta)^2-3tan^4\theta+3sec^2\theta tan^2\theta`
`=(sec^2\theta-tan^2\theta)^2+3tan^2\theta(sec^2\theta-tan^2\theta)`
`Use\text{ }sec^2\theta-tan^2\theta=1.`
`=1^2+3tan^2\theta(1)`
`=1+3tan^2\theta`
`Use\text{ }tan^2\theta=sec^2\theta-1.`
`=1+3(sec^2\theta-1)`
`=1+3sec^2\theta-3`
`=3sec^2\theta-2`

We get the right-hand side of the equation.

Final answer:

`sec^4\theta+sec^2\theta tan^2\theta-2tan^4\theta=3sec^2\theta-2`