## meyruhstfu94 Group Title Find in the missing step: sec^4 x -2 sec^2 x+1=(sec^2x-1)^2 =????????? =tan^4 x 2 years ago 2 years ago

(tan^2x)^2

[ sec^2x-1= tan^2x ]

3. meyruhstfu94 Group Title

how is that?

4. meyruhstfu94 Group Title

is it because tan^2x=sec^2x-1?

Yes

its an identity =)

7. meyruhstfu94 Group Title

okay thanks

8. darthsid Group Title

So, the best way to do this is to go back to basics of a right triangle!! |dw:1321912053720:dw| H = hypotenuse, B = base, P = perpendicular $\sec(x) = \frac{H}{B}$ $\sec^2(x) = \frac{H^2}{B^2}$ $Sec^2(x) - 1 = \frac{H^2}{B^2} - 1 = \frac{H^2 - B^2}{B^2}$ Now, according to the Pythagoras theorem, $H^2 = P^2 + B^2$ Which means $H^2 - B^2 = P^2$ So, $Sec^2(x) - 1 = \frac{H^2 - B^2}{B^2} = \frac{P^2}{B^2}$ $\frac{P}{B} = tan(x)$ Thus, you get your answer :)