If A - B and C represents the angles of a triangle- then cos A -cos B -cos C -1 equal toA- 4 sinA-2sinB-2sinC-2B- 4 cosA-2cosB-2cosC-2C- -4 sinA-2sinB-2sinC-2D- -4 cos2-2cos B -2cos C -2

The correct option is A 4sinA2sinB2sinC2( cos A + cos B )+ cos C 1 =2cosA+B2cosA B2 +cos C 1 =2cosπ C2cosA B2 +cos C 1 =2sinC2cosA B2 + 1 2sin^2 C2 ...

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Byju's Answer

Standard XII

Mathematics

Conditional Identities

If A , B and ...

Question

If $A, B$ and $C$ represents the angles of a triangle, then $cos A + cos B + cos C - 1$ equal to

- 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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- 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

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4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

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Solution

The correct option is B $4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$
$(cos A + cos B) + cos C - 1 = 2 cos \frac{A + B}{2} cos \frac{A - B}{2} + cos C - 1 = 2 cos \frac{π - C}{2} cos \frac{A - B}{2} + cos C - 1 = 2 sin \frac{C}{2} cos \frac{A - B}{2} + 1 - 2 {sin}^{2} \frac{C}{2} - 1 = 2 sin \frac{C}{2} [cos \frac{A - B}{2} - sin \frac{C}{2}] = 2 sin \frac{C}{2} [cos \frac{A - B}{2} - sin \frac{π - (A + B)}{2}] = 2 sin \frac{C}{2} [cos \frac{A - B}{2} - cos \frac{A + B}{2}] = 2 sin \frac{C}{2} [2 sin \frac{A}{2} sin \frac{B}{2}] = 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$

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Similar questions

Q. Prove

\frac{4 sinAsinBsinC}{4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}} = 8 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2} .

Q. Assertion :If in a triangle ,

cos A + 2 cos B + cos C = 2,

then

a, b, c

must be in A.P Reason:

cos A + cos B + cos C = 1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

Q.
lf

A + B + C = 180^{\circ}

. Match the following
List -I List -II

1.

cos A + cos B + cos C

2 + 2 cos A cos B cos C

{sin}^{2} A + {sin}^{2} B + {sin}^{2} C

4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

sin 2 A + sin 2 B + sin 2 C

1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

sin A + sin B + sin C

4 sin A sin B sin C

Q. In a

Δ A B C

prove that

cosA + cos B + cos C = 1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

Q. In any triangles ABC prove the identities.

cos A + cos B + cos C = 1 + 4 sin (A / 2) sin (B / 2) sin (C / 2)

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If A,B and C represents the angles of a triangle, then cosA+cosB+cosC−1 equal to

The correct option is B 4sinA2sinB2sinC2(cosA+cosB)+cosC−1=2cosA+B2cosA−B2+cosC−1=2cosπ−C2cosA−B2+cosC−1=2sinC2cosA−B2+1−2sin2C2−1=2sinC2[cosA−B2−sinC2]=2sinC2[cosA−B2−sinπ−(A+B)2]=2sinC2[cosA−B2−cosA+B2]=2sinC2[2sinA2sinB2]=4sinA2sinB2sinC2

If $A, B$ and $C$ represents the angles of a triangle, then $cos A + cos B + cos C - 1$ equal to