cosA+cos(240°+A)+cos(240°-A)=?
cosA
0
3sinA
3cosA
Explanation for the correct option:
Given, cosA+cos(240°+A)+cos(240°–A)
Using the identity cos(A+B)+cos(A–B)=2cosAcosB, we get
=cosA+2cos240°cosA
=cosA[1+2cos(270°–30°)]
=cosA[1+2(-sin30°)] ; ∵cos(3π2-θ)=-sinθ
=cosA1–2×12
=cosA[1–1]
=0
Hence, Option ‘B’ is Correct.
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