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question mark

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

Sid , 9 Years ago
Grade 12th pass
anser 5 Answers
JOEL JOHNSON

Last Activity: 9 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

Last Activity: 8 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

Last Activity: 6 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

Last Activity: 6 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

Last Activity: 4 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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