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Question

In triangle ABC,sinA+sinB+sinCsinA+sinBsinC is equal to
  1. tanA2cotB2
  2. cotA2cotB2
  3. tanA2tanB2
  4. cotA2tanB2


Solution

The correct option is B 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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