anonymous
  • anonymous
tan^x-sec^4x = 1 - 2sec^2x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
...is this supposed to be \[\tan(x) - \sec^4(x) = 1-2\sec^2(x)\]?
anonymous
  • anonymous
oh no. oops. its tan^4(x) - sec^4(x) = 1 - 2 sec^2(x)
myininaya
  • myininaya
ok you have the difference of squares here a^2+b^2=(a-b)(a+b)

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anonymous
  • anonymous
okay so that would make it (tan^2x-sec^2x) (tan^2x+sec^2x)
myininaya
  • myininaya
so we have (tan^2x)^2-(sec^2x)^2=[(tanx)^2+(secx)^2][(tanx)^2-(secx)^2]
myininaya
  • myininaya
yes and 1=(secx)^2-(tanx)^2
myininaya
  • myininaya
so we have -((tanx)^2+(sec^2))((secx)^2-(tanx)^2)
myininaya
  • myininaya
-((tanx)^2+(secx)^2)(1)
myininaya
  • myininaya
ok the other side is in terms of sec so put this side in terms of sec
anonymous
  • anonymous
ok thanks
myininaya
  • myininaya
you got it?
myininaya
  • myininaya
this problem is basically done
myininaya
  • myininaya
so we have -((secx)^2-1+(secx)^2)=-(2(secx)^2-1)=1-2(secx)^2
myininaya
  • myininaya
which is the other side of the equation

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