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Find in the missing step: sec^4 x -2 sec^2 x+1=(sec^2x-1)^2 =????????? =tan^4 x

Mathematics
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(tan^2x)^2
[ sec^2x-1= tan^2x ]
how is that?

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Other answers:

is it because tan^2x=sec^2x-1?
Yes
its an identity =)
okay thanks
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 :)

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