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Question

In a triangle ABC,the value of sinA+sinB+sinC is :


A

4sinA2sinB2sinC2

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B

4cosA2cosB2cosC2

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C

4cosA2sinB2sinC2

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D

4cosA2sinB2cosC2

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Solution

The correct option is B

4cosA2cosB2cosC2


Finding the value of sinA+sinB+sinC in a triangle.

In any triangle, the sum of all the interior angles is always 180°.

Therefore, In the triangle ABC, A+B+C=180°

Now, solving for sinA+sinB+sinC:

sinA+sinB+sinC=2sinA+B2cosA-B2+sinC=2sinA+B2cosA-B2+2sinC2cosC2=2sin180°-C2cosA-B2+2sinC2cosC2=2cosC2cosA-B2+2sinC2cosC2=2cosC2cosA-B2+sinC2=2cosC2cosA-B2+sin180°-(A+B)2=2cosC2cosA-B2+cosA+B2=4cosC2cosB2cosA2sinA+sinB+sinC=4cosA2cosB2cosC2

Hence, Option (B) is correct.


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