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Question

If a less or equal than fraction numerator s e c to the power of 2 end exponent invisible function application theta minus t a n invisible function application theta over denominator s e c to the power of 2 end exponent invisible function application theta plus t a n invisible function application theta end fraction less or equal than b then a+b=

  1. fraction numerator 1 over denominator 3 end fraction    
  2. fraction numerator 5 over denominator 3 end fraction    
  3. fraction numerator 10 over denominator 3 end fraction    
  4. fraction numerator 8 over denominator 3 end fraction    

The correct answer is: fraction numerator 10 over denominator 3 end fraction


    t a n invisible function application theta equals x
    fraction numerator 1 minus x plus x to the power of 2 end exponent over denominator 1 plus x plus x to the power of 2 end exponent end fraction equals y
    x to the power of 2 end exponent left parenthesis 1 minus y right parenthesis minus x left parenthesis 1 plus y right parenthesis plus left parenthesis 1 minus y right parenthesis equals 0
    left parenthesis 1 plus y right parenthesis to the power of 2 end exponent minus 4 left parenthesis 1 minus y right parenthesis to the power of 2 end exponent greater or equal than 0
    3 y to the power of 2 end exponent minus 10 y plus 3 less or equal than 0
    y element of open square brackets fraction numerator 1 over denominator 3 end fraction comma 3 close square brackets
    a equals fraction numerator 1 over denominator 3 end fraction comma b equals 3 semicolon a plus b equals fraction numerator 10 over denominator 3 end fraction

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