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if A+B+C=180 then 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

if A+B+C=180
then 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

Grade:12th pass

5 Answers

JOEL JOHNSON
35 Points
6 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
4 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
3 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
3 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
askIITians Faculty 614 Points
one year ago
dear student in objective exam u can put the value of angles randomly or u can solve as
 
 
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

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