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Let fn(Theta)=(tan(Theta/2))(1 +sec(Thet...

Let` f_n(Theta)=(tan(Theta/2))(1 +sec(Theta))(1+sec(2Theta)).....(1+sec(2^nTheta)`.. then

A

`f_(2) ((pi)/(16))=1`

B

`f_(3) ((pi)/(32)) =1`

C

`f_(4) ((pi)/(64)) =1`

D

`f_(5) ((pi)/(128))=-1`

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The correct Answer is:
D
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f_(n)(Theta)=(tan((Theta)/(2)))(1+sec(Theta))(1+sec(2Theta))......(1+sec(2^(n)Theta)

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For positive integer n if f_(n)(theta) = tan ""(theta)/(2) (1 + sec theta) (1+ sec2 theta) (1 + sec4 theta)"…."( 1 + sec2^(n)theta) then find the value of f_(2) ((pi)/(16)) " and " f_(5) ((pi)/(128)) .

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