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Question

In ΔABC if cotA2.cotB2=c;cotB2cotC2=a; and cotC2cotA2=b then 1sa+1sb+1sc=

A
-1
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B
0
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C
1
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D
2
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Solution

The correct option is D 2
tanA/2=Δ(sa)s
stanA/2Δ=1(sa)
SΔ(tanA/2+tanB/2+tanC/2)
cotA/2=(sa)sΔ,cotB/2=(sb)sΔ,cotC/2=(sc)sΔ
(sc)(sc)×(sa)sΔ×(sb)sΔ=C
s(sc)=C
1(sc)=cs, ---(i) Similarly1(sb)=bs,...(ii)1(sa)=as....(iii)
after observing eq(i),eq(ii) and eq(iii)
So the equation asked in the qs is equal to cs+bs+as
=a+b+cs
=2.

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