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If u equals logtan invisible function application open parentheses fraction numerator pi over denominator 4 end fraction plus fraction numerator x over denominator 2 end fraction close parentheses comma then cos h u is equal to

  1. sec invisible function application x    
  2. text cosec end text x    
  3. tan invisible function application x    
  4. sin invisible function application x    

The correct answer is: sec invisible function application x


    u equals logtan invisible function application open parentheses fraction numerator pi over denominator 4 end fraction plus fraction numerator x over denominator 2 end fraction close parentheses
    Þ fraction numerator e to the power of u end exponent over denominator 1 end fraction equals fraction numerator 1 plus tan invisible function application fraction numerator x over denominator 2 end fraction over denominator 1 minus tan invisible function application fraction numerator x over denominator 2 end fraction end fraction
    cos h invisible function application u equals fraction numerator e to the power of u end exponent plus e to the power of negative u end exponent over denominator 2 end fraction equals fraction numerator e to the power of 2 u end exponent plus 1 over denominator 2 e to the power of u end exponent end fraction equals fraction numerator open parentheses fraction numerator 1 plus tan invisible function application fraction numerator x over denominator 2 end fraction over denominator 1 minus tan invisible function application fraction numerator x over denominator 2 end fraction end fraction close parentheses to the power of 2 end exponent plus 1 over denominator 2. open parentheses fraction numerator 1 plus tan invisible function application fraction numerator x over denominator 2 end fraction over denominator 1 minus tan invisible function application fraction numerator x over denominator 2 end fraction end fraction close parentheses end fraction
    equals fraction numerator 2 open parentheses 1 plus tan to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction close parentheses over denominator 2 open parentheses 1 minus tan invisible function application fraction numerator x over denominator 2 end fraction close parentheses open parentheses 1 plus tan invisible function application fraction numerator x over denominator 2 end fraction close parentheses end fraction equals fraction numerator 1 plus tan to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction over denominator 1 minus tan to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction end fraction
    equals fraction numerator 1 over denominator fraction numerator 1 minus tan to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction over denominator 1 plus tan to the power of 2 end exponent invisible function application fraction numerator x over denominator 2 end fraction end fraction end fraction equals fraction numerator 1 over denominator cos invisible function application x end fraction equals sec invisible function application x.

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