-cos(A+B)cos(B-A) =...

`-cos(A+B)cos(B-A) =`

A

`cos^(2)A - sin^(2)B`

B

`cos^(2)B - sin^(2)A`

C

`sin^(2)B - cos^(2)A`

D

None of these

Verified by Experts

The correct Answer is:

A

DEFINITE INTEGRATION & ITS APPLICATION
RESONANCE ENGLISH|Exercise High Level Problem|26 Videos
EQUATIONS
RESONANCE ENGLISH|Exercise EXERCISE-2 (PART-II: PREVIOUSLY ASKED QUESTION OF RMO) |5 Videos

Similar Questions

Explore conceptually related problems

Prove that: (sin(A+B)+"sin"(A-B))/(cos(A+B)+"cos"(A-B))=t a n A

|(1,cos(A-B), cos(A-C)),(cos(B-A), 1, cos(B-C)),(cos(C-A), cos(C-B), 1)| (A) 0 (B) 1 (C) cosAcosBcosC (D) none of these

Prove that: a) sin(A+B+C) + sin(A-B-C)+sin(A+B-C) + sin(A-B+C) = 4sinAcosBcosC b) cos(A+B+C)+cos(A+B-C)+cos(B+C-A)+cos(C+A-B)=4cosAcosBcosC

If B be the exterior angle of a regular polygon of n - sides and A is any constant then cos A + cos(A+ B)+cos(A+2B)....n terms is equal to

Prove that: cos(A+B+C)+cos(A-B+C)+cos(A+B-C)+cos(-A+B+C)=4cos A\ cos B\ cos C

If A+B+C=180^(@), " then " (cosAcosC+cos(A+B)cos(B+C))/(cosAsinC-sin(A+B)cos(B+C)) simplifies to :

Prove that |{:(cos(A-P),cos(A-Q),cos(A-R)),(cos(B-P),cos(B-Q),cos(B-R)),(cos(C-P),cos(C-Q),cos(C-R)):}| =0.

For all values of A ,B ,Ca n dP ,Q ,R show that |"cos"(A-P)"cos"(A-Q)"cos"(A-R)"cos"(B-P)"cos"(B-Q)"cos"(B-R)"cos"(C-P)"cos"(C-Q)"cos"(C-R)|=0

Prove that: (cos(A+B+C)+cos(-A+B+C)+cos(A-B+C)+"cos"(A+B-C))/(sin(A+B+C)+sin(-A+B+C)+sin(A-B+C)-"sin"(A+B-C))=cot C

Prove that: i) sin(A+B)cos(A-B)-cos(A+B)sin(A-B)=sin2B ii) cos(45^(@)-A)cos(45^(@)-B)-sin(45^(@)-A)sin(45^(@)-B)=sin(A+B)