For a positive integer n- let f n - - tan- 2 1-sec- 1-sec2- 1-sec4- - 1-sec 2 n - -Then

The correct options are A f _ 2 ( π 16 ) =1 B f _ 4 ( π 64 ) =1 C f _ 3 ( π 32 ) =1 D f _ 5 ( π 128 ) =1 fx(θ)=tanθ2(1+θ)(1+^2θ)(1+^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of Cosines of Angles in Arithmetic Progression

For a positiv...

Question

For a positive integer n, let $f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . . . (1 + s e c 2^{n} θ) .$ Then

f_{2} (\frac{π}{16}) = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f_{4} (\frac{π}{64}) = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f_{3} (\frac{π}{32}) = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f_{5} (\frac{π}{128}) = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $f_{2} (\frac{π}{16}) = 1$
B $f_{4} (\frac{π}{64}) = 1$
C $f_{3} (\frac{π}{32}) = 1$
D $f_{5} (\frac{π}{128}) = 1$
$f x (θ) = tan \frac{θ}{2} (1 + sec θ) (1 + {sec}^{2} θ) (1 + {sec}^{2} 2 θ) \dots \dots \dots (1 + sec 2^{n} θ)$
$= tan \frac{θ}{2} (1 + \frac{1}{cos θ}) (1 + sec 2 θ) \dots \dots (1 + sec 2^{n} θ)$
$= tan \frac{θ}{2} (1 + \frac{1 + {tan}^{2} θ / 2}{1 - {tan}^{2} θ / 2}) (1 + \frac{1}{cos 2 θ}) \dots . . (1 + sec 2^{n} θ)$
$= tan \frac{θ}{2} (\frac{2}{1 - {tan}^{2} θ / 2} (1 + \frac{1 + {tan}^{2} θ}{1 - {tan}^{2} θ}) \dots . (1 + sec 2^{n} θ)$
$= tan 2 θ (1 + {sec}^{2} 2 θ) \dots . . (1 + sec 2^{n} θ)$
$= tan 2 θ (1 + \frac{1 - {tan}^{2} 2 θ}{1 + {tan}^{2} 2 θ}) \dots \dots . (1 + sec 2^{n} θ)$
$= tan 2^{2} θ \dots \dots \dots . . (1 + sec 2^{n} θ)$
$= tan 2^{n} θ$
$∴ f x (θ) = tan 2^{n} θ$
(i) $f_{2} (π / 16) = tan 2^{2} π / 16 = tan \frac{4 π}{16} = tan \frac{π}{4} = 1$
(ii) $f_{4} (\frac{π}{64}) = tan (\frac{2^{4} π}{64}) = tan (\frac{16 π}{64}) = tan \frac{π}{4} = 1$
(iii) $f_{3} (\frac{π}{32}) = tan (\frac{2^{3} π}{32}) = tan (\frac{8 π}{32}) = tan \frac{π}{4} = 1$
(iv) $f_{5} (\frac{π}{128}) = tan (\frac{2^{5} π}{128}) = tan (\frac{32 π}{128}) = tan \frac{π}{4} = 1$ .

Suggest Corrections

Similar questions

For a positive integer n, let
$f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) \dots (1 + s e c 2^{n} θ)$ .
Then

For a positive integer n, let $f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . . . . (1 + s e c 2^{n} θ) . T h e n$

Q. For a positive integer

n,

let

f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) . . . (1 + sec 2^{n} θ) .

Then

Q. For a positive integers n, let

f_{n} (θ) = t a n (θ / 2) (1 + s e c θ) (1 + s e c 2 θ) . . . (1 + s e c 2^{n} θ)

then

Q. Let

f_{n} (θ) = t a n \frac{θ}{2} . (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . (1 + s e c 2^{''} θ) .

Then

Join BYJU'S Learning Program

For a positive integer n, let fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ).....(1+sec2nθ).Then

For a positive integer n, let $f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . . . (1 + s e c 2^{n} θ) .$ Then