anonymous
  • anonymous
square root 1+tan^2x/1-sin^2x all is under the square root. i did sec^2theta/cos^2theta whats next?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
sec^4 2x
anonymous
  • anonymous
you just need this trigonometric function to be solved or do you have anything to be proved?
anonymous
  • anonymous
find what it is equivalent to

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anonymous
  • anonymous
then that's the answer! :)
anonymous
  • anonymous
\[\sqrt{1+\tan^2x}\div 1-\sin^2x\]
anonymous
  • anonymous
both of these r under the square root
anonymous
  • anonymous
ahh sorry i didn't notice that!then the answer is sec^2 x
anonymous
  • anonymous
how did u get that?
anonymous
  • anonymous
1+tan^2 x= sec^2 x (that's trigonometry theory) sin^x + cos^x=1 (that's trigonometry theory) therefore [1-sin ^2 x= cos ^2 x] so √1+tan2x ÷ √1−sin2x = √sec^2 x ÷ √cos ^2 x =sec x ÷ cos x =sec ^2 x (because 1÷cos x =sec x)
anonymous
  • anonymous
oh i c i understand up to the 4th line and i got the answer on my own but up to the fourth line .wcan u explain line 5 and 6 for me .how did u cancel it and set up the fraction,.
anonymous
  • anonymous
=sec x ÷ cos x = sec x * 1/ cos x = sec x * sec x ( because 1/cos x = sec x) =sec ^2 x
anonymous
  • anonymous
k thanks so much and thanks for your patience .

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