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.

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:

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