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meyruhstfu94
Group Title
Find in the missing step:
sec^4 x 2 sec^2 x+1=(sec^2x1)^2
=?????????
=tan^4 x
 3 years ago
 3 years ago
meyruhstfu94 Group Title
Find in the missing step: sec^4 x 2 sec^2 x+1=(sec^2x1)^2 =????????? =tan^4 x
 3 years ago
 3 years ago

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AditiMeow Group TitleBest ResponseYou've already chosen the best response.2
(tan^2x)^2
 3 years ago

AditiMeow Group TitleBest ResponseYou've already chosen the best response.2
[ sec^2x1= tan^2x ]
 3 years ago

meyruhstfu94 Group TitleBest ResponseYou've already chosen the best response.0
how is that?
 3 years ago

meyruhstfu94 Group TitleBest ResponseYou've already chosen the best response.0
is it because tan^2x=sec^2x1?
 3 years ago

AditiMeow Group TitleBest ResponseYou've already chosen the best response.2
its an identity =)
 3 years ago

meyruhstfu94 Group TitleBest ResponseYou've already chosen the best response.0
okay thanks
 3 years ago

darthsid Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 3 years ago
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