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Question

In triangle ABC,sinA+sinB+sinCsinA+sinBsinC is equal to

A
tanA2cotB2
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B
cotA2tanB2
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C
cotA2cotB2
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D
tanA2tanB2
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Solution

The correct option is C cotA2cotB2
We know that
A+B+C=π
Now,
sinA+sinBsinC=2sinA+B2cosAB22sinC2cosC2(sin(A+B2)=sin(π2C2)=cosC2)=2cosC2[cosAB2sinC2]=2cosC2[cosAB2cosA+B2](sin(C2)=sin(π2A+B2)=cosA+B2)=4sinA2sinB2cosC2

Similarly
sinA+sinB+sinC=2cosC2[cosAB2+cosA+B2]=4cosA2cosB2cosC2

sinA+sinB+sinCsinA+sinBsinC=cotA2cotB2

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