Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

lf cos α+cos β+cos γ=0=sin α+sin β+sin γ  then xsin2α-β-γ.xsin2β-γ-α.xsin2γ-α-β=

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

x

b

1

c

0

d

x3

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Let  a=cis α,  b=cis β,  c=cis γ 

a+b+c=(cos α+i sin α)+(cos β+i sin β)+(cos γ+i sin γ)

=(cos α+cos β+cos γ)+i(sin α+sin β+sin γ)

a+b+c=0    a3+b3+c3=3abc

a3abc+b3abc+c3abc=3    a2bc+b2ca+c2ab=3

=(cis α)2cis β cis γ+(cis β)2cis γ cis α+(cis γ)2cis α cis β=3

cis (2α-β-γ)+cis(2β-γ-α)+cis(2γ-α-β)=3

cos(2α-β-γ)=3,  sin(2α-β-γ)=0

xsin(2α-β-γ)·xsin(2β-γ-α)·x(2γ-α-β) =x sin(2α-β-γ) =x0=1

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring
lf cos α+cos β+cos γ=0=sin α+sin β+sin γ  then xsin2α-β-γ.xsin2β-γ-α.xsin2γ-α-β=