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meyruhstfu94

  • 4 years ago

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

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  1. Aditi-Meow
    • 4 years ago
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    (tan^2x)^2

  2. Aditi-Meow
    • 4 years ago
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    [ sec^2x-1= tan^2x ]

  3. meyruhstfu94
    • 4 years ago
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    how is that?

  4. meyruhstfu94
    • 4 years ago
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    is it because tan^2x=sec^2x-1?

  5. Aditi-Meow
    • 4 years ago
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    Yes

  6. Aditi-Meow
    • 4 years ago
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    its an identity =)

  7. meyruhstfu94
    • 4 years ago
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    okay thanks

  8. darthsid
    • 4 years ago
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    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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