#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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. Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
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!!!

Regards,