For a positive integer n , let f'n'(a)=(tan a/2)(1+sec a)(1+sec 2a)(1+ sec 4a)....(1+sec 2^nA) ..then..1. f'2'(pi/16)=12.f'3'(pi/32)=13.f'4'(pi/64)=14.f'5'(pi/128)=1The numbers under quotes are subscripts. Thank you.

Dear Balaram,

Solution:- f n'(a) = (tan a/2)(1+ sec a)(1+sec 2a)(1+sec 4a).......(1+sec 2ⁿa)

Consider, first 2 terms- (tan a/2) (1+ sec a) = (tan a/2) (1+ cos a)/ (cos a)

                                         = [(sin a/2) (2cos² a/2)]/ [(cos a/2)(cos a)] 

                                         = [(sin a/2) (2cos a/2)]/ [(cos a)] = (sin a)/ (cos a) = tan a

So,  f n'(a) = (tan a)(1+sec 2a)(1+sec 4a).......(1+sec 2ⁿa)

On doing same exercise for first two terms till the last term, we get-

f n'(a) = (tan  2^n-1a)(1+sec 2ⁿa) = tan 2ⁿa

Therefore, f 2'(∏/16) = tan[2²(∏/16)] = tan(∏/4) = 1

f 3'(∏/32) = tan[2³(∏/32)] = tan(∏/4) = 1

f 4'(∏/64) = tan[2⁴(∏/64)] = tan(∏/4) = 1

f 5'(∏/132) = tan[2⁵(∏/132)] = tan(∏/4) = 1

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