wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For a positive integer n, let fn(θ)=(tanθ2)(1+sec θ)(1+sec 2θ)(1+sec 4θ)......(1+sec 2nθ). Then


A

f2(π16)=1

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

f3(π32)=1

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

f4(π64)=1

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

f5(π128)=1

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

The correct option is D

f5(π128)=1


(a,b,c,d) fn(θ)=sin(θ2)cos(θ2)[2 cos2 θ2cos θ.2 cos2 θcos 2θ.2 cos2 θcos 4θ..]

Combine first two factors, fn(θ)=sin θcos θ[2 cos2 θcos 2θ32 cos2 2θcos 4θ...]

Again combine first two factors,

fn(θ)=tan 2θ[2 cos2 2θcos 4θ...]=tan(2nθ) f2(π16)=tan4π16=tan(π4)=1f3(π32)=tan8π32=tan(π4)=1

f4(π64)=tan16π64=tan(π4)=1f5(π128)=tan 32π128=tan(π4)=1


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Multiple and Sub Multiple Angles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon