# if A+B+C=180then value of  cotA + cotB + cotC  is                        cotA.cotB.cotC     a.) 1          b.) cotA.cotB.cotC.     c.) -1          d.) 0  this question is asked in bitsat 2012 question paper (Q.112)  given on your site

JOEL JOHNSON
35 Points
8 years ago
here 1800= pi
now A+B+C=pi
A/2 + B/2= pi/2 – C/2
cot[A/2+B/2]=cot[pi/2 - C/2]
cotA/2.cotB/2 – 1/ cotA/2 + cotB/2= tanC/2= 1/ cotC/2
therefore, cotA/2.cotB/2.cotC/2. =cotA/2 + cotB/2 +cotC/2
this implies that  cotA/2 + cotB/2 + cotC/2 / cotA/2.cotB/2.cotC/2= 1
so the answer is a) 1

Esha
17 Points
6 years ago
A better way to solve this question:Expand tan(A+B+C) as{tanA+tan(B+C)}/1-tan(A)(B+C) Put tan(B+C)=(tanB+tanC)/1-tanBtanCAnd then saying that Denominator of resultant!=0, equate Numerator=0 because tan(A+B+C)=tan(pi)=0
Akshay Jyothis
24 Points
5 years ago
But substituting A=B=C=60 Deg .

$cot 60 = \frac{1}{\sqrt{3}} \therefore \frac{3 cot 60}{cot 60^3}= \frac{\frac{3}{\sqrt{3}}}{\frac{1}{\sqrt{3}^3}} =9$

Doesn’t gives an answer from the options.

I personally recommend substituting values satisfying the given conditions for such questions.Proving the equation kills your time.
Prateek Gaur
13 Points
5 years ago
tan(A+B+C)=tanA+tanB+tanC-tanAtanBtanC/tanAtanB-tanBtanC-tanCtanA
Solve you will get ur answer
If asked for cot do the resiprocal and convert into tan it will be easier to solve

ankit singh