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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
7 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
5 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
4 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
4 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
2 years 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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