sin(B+A)+cos(B-A)sin(B-A)+cos(B+A)=
(cosB+sinB)(cosB–sinB)
(cosA+sinA)(cosA–sinA)
(cosA–sinA)(cosA+sinA)
None of these
Explanation for the correct option:
Simplifying the equations using trigonometric identities:
sin(B+A)+cos(B-A)sin(B-A)+cos(B+A)
Using the formulas,
sin(x+y)=cosxsiny+sinxcosysin(x-y)=cosxsiny-sinxcosycos(x+y)=cosxcosy+sinxsinycos(x-y)=cosxcosy-sinxsiny
=sinBcosA+cosBsinA+cosBcosA+sinBsinAsinBcosA–cosBsinA+cosBcosA–sinBsinA[sin(x+y)=cosxsiny+sinxcosy;sin(x-y)=cosxsiny-sinxcosy;cos(x+y)=cosxcosy+sinxsiny;cos(x-y)=cosxcosy-sinxsiny]=cosAsinB+cosB+sinAcosB+sinBcosA(sinB+cosB)–sinA(cosB+sinB)=(sinB+cosB)(cosA+sinA)(sinB+cosB)(cosA–sinA)=cosA+sinAcosA-sinA
Therefore, the correct answer is option (B).
For any two sets A and B, prove that
(i) (A∪B)−B=A−B
(ii) A−(A∩B)=A−B
(iii) A−(A−B)=A∩B
(iv) A∪(B−A)=A∪B
(v) (A−B)∪(A∩B)=A
If a1,a2,a3,a4 are in AP, then