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For +ve integer n, letfn(theta)=tan(thet...

For +ve integer n, let`f_n(theta)=tan(theta/2)(1+sectheta)(1+sec2theta)(1+sec4theta)…..(1+sec2^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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PATHFINDER-TRIGONOMETRIC RATIOS AND IDENTITIES-QUESTION BANK
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