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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)=1 2.f'3'(pi/32)=1 3.f'4'(pi/64)=1 4.f'5'(pi/128)=1 The numbers under quotes are subscripts. Thank you.

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)=1


2.f'3'(pi/32)=1


3.f'4'(pi/64)=1


4.f'5'(pi/128)=1


 


The numbers under quotes are subscripts. Thank you.

Grade:12

1 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
12 years ago

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  2n-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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All the best Balaram!!!

Regards,
Askiitians Experts
Priyansh Bajaj

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