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Question

If A(theta) = open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets and AB = I, then (sec2 theta) B is equal to-

  1. A(theta)    
  2. A(–theta)    
  3. A(theta/2)    
  4. A(–theta/2)    

The correct answer is: A(–theta)


    As AB = I, we get B = A–1
    therefore B equals fraction numerator 1 over denominator 1 plus tan squared invisible function application theta end fraction open square brackets table row 1 cell negative tan invisible function application theta end cell row cell tan invisible function application theta end cell 1 end table close square brackets
    rightwards double arrow(sec2theta) B = open square brackets table row 1 cell negative tan invisible function application theta end cell row cell tan invisible function application theta end cell 1 end table close square brackets equals A left parenthesis negative theta right parenthesis

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