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Question

If A + B + C = π. Then

Column AColumn B1.sin2A + sin2B + sin2CA.cosA2 cos B2 cos C22.sinA + sinB + sinCB.1+4sinA2 sinB2 sinC23.cos2A + cos2B + cos2CC.4sinA sinB sinC4.cosA + cosB + cosCD.-1-4cosA + cosB + cosC


A

1 - B, 2 - D, 3 - C, 4 - A

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B

1 - A, 2 - C, 3 - D, 4 - B

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C

1 - C, 2 - A, 3 - D, 4 - B

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D

1 - C, 2 - A, 3 - B, 4 - D

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Solution

The correct option is C

1 - C, 2 - A, 3 - D, 4 - B


1. sin2A + sin2B + sin2C

we need to just simplify the given expression.

We know,

sinA + sinB = 2 sin(A+B)2. cos(AB)2

2sin(2A+2B)2.cos(2A+2B)2 + sin 2[π - (A+B)]

2 sin(A+B).cos(A-B)+ sin (2π - 2(A+B))

2 sin(A+B).cos(A-B) - sin 2(A+B)

2 sin(A+B).cos(A-B) - 2 sin(A+B). cos(A+B) {sin 2A = 2 sinA cosA}

2 sin(A+B){cos(A-B) - cos(A+B)}

= 2 sin(π - C) [2sin(AB+A+B)2sin(A+BA+B)2]

= 2 sinC{2 . sinA . sinB}

= 4 sinA sinB sinC

2.sinA sinB sinC

2 sin(A+B)2 . cos(AB)2 + sin (π - (A+B)) {C = π - (A+B)}

= 2 sin(A+B)2 . cos(AB)2 + sin(A+B)

= 2 sin(A+B)2 . cos(AB)2 + 2 sin(A+B)2 . cos(A+B)2 [sinA=2sinA2.cosA2]

= 2 sin(A+B)2 [cosAB2+cosA+B2]

= 2sin(πC2)×[2cos(AB+A+B2)2×cos(A+BA+B2)2]

= 2 sin (π2C2).2 cosA2 . cosB2

= 2 cosC2 . 2cosA2 . cosB2

= 4 cosA2 cosB2 cosC2

3. cos2A + cos2B + cos2C

2 cos (2A+2B)2 cos(2A2B)2 + cos 2(π - (A+B))

= 2 cos(A+B). cos(A-B) + {cos (2π - 2(A+B))}

= 2 cos(A+B). cos(A-B) + cos 2(A+B)

= 2 cos(A+B). cos(A-B) + 2cos2(A+B) - 1 {cos 2A = 2cos2 A - 1}

= 2 cos(A+B) {cos(A-B) + cos(A+B)} - 1

= 2cos(πC){2cos(AB+A+B)2cos(A+BA+B)2}1

= -2cos C {2 cosA cosB} - 1

= -4 cosA cosB cosC - 1

4. cosA + cosB + cosC

= 2 cos (A+B)2 . cos (AB)2 + cos{π - (A+B)}

= 2 cos (A+B)2 . cos (AB)2 +{-cos (A+B)}

= 2 cos (A+B)2 . cos (AB)2 - cos(A+B)

= 2 cos (A+B)2 . cos (AB)2 - [2cos2(A+B)21] {cos 2A = 2 cos2 A - 1}

= 2 cos (A+B)2 . cos (AB)2 - 2cos2(A+B)2+1

= 2 cos (A+B)2 [cos(AB)2cos(A+B)2] + 1

= 2 cos (π+C)2 [2sinA2.sinB2+1]

= 2 sinC2 . 2 sinA2 . sinB2 + 1

= 1 + 4sinA2 sinB2 sinC2


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