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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





The numbers under quotes are subscripts. Thank you.

5 years ago


Answers : (1)


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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Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.

All the best Balaram!!!

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Priyansh Bajaj

5 years ago

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