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If cos(A+B)sin(C-D)=cos(A-B)sin(C+D) then show that tanA tanB tanC+tanD=0

If cos(A+B)sin(C-D)=cos(A-B)sin(C+D) then show that tanA tanB tanC+tanD=0

Grade:11

2 Answers

Arun
25750 Points
5 years ago

cos(a+b) sin(c-d) = cos(a-b) sin(c+d)

( cosAcosB - sinAsinB) (sincCosD- cosC sinD) = ( cosAcosB + sinAsinB) (sincCosD+ cosC sinD)

- cosA cosBcosCsinD - sinA sinB sinC cosD = cosA cosB cosC sinD + sinA sinB sinC cosD

2 cosA cosBcosCsinD + 2sinA sinB sinC cosD=0

sinA sinB sinC cosD = - cosA cosBcosCsinD

tanA tanB tanC = - tanD

tannA tanB tanC + tanD =0

 

 

Regards

Arun (askIITians forum expert)

Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

cos(a+b) sin(c-d) = cos(a-b) sin(c+d)
( cosAcosB - sinAsinB) (sincCosD- cosC sinD) = ( cosAcosB + sinAsinB) (sincCosD+ cosC sinD)
- cosA cosBcosCsinD - sinA sinB sinC cosD = cosA cosB cosC sinD + sinA sinB sinC cosD
2 cosA cosBcosCsinD + 2sinA sinB sinC cosD=0
sinA sinB sinC cosD = - cosA cosBcosCsinD
tanA tanB tanC = - tanD
tannA tanB tanC + tanD =0

Thanks and Regards

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