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Question

tan1abc(a+b+c)+tan1bac(a+b+c)+tan1cac(a+b+c)=

A
\N
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B
π
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C
π
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D
None
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Solution

The correct option is B π
θ=tan1ar+tan1br+tan1cr
Where r={(a+b+c)abc}
=π+tan1ar+br1arbr+tan1cr1abr2=1(a+b+c)c[a,b,c>0]=(a+b)c<0=π+Tan1(cr)+tan1cr=πtan1cr+tan1cr=π

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