Question

5. Let ABC be a triangle such that cotA+cotB+ cotC= _3 then prove that ABC is an
equilateral triangle.​

Answer 1

Answer:

if i'm not wrong,

Ques should be given like:

if cotA+cotB+cotC = √3, then P.T ....

Step by step Explanation:

We have, cotA+cotB+cotC=√3

Squaring, 

cot²A+cot²B+cot²C+2cotA.cotC

+2cotB.cotC+2cotC.cotA = 3 .....(1)

Now in △ABC. A+B+C=π ⇒ A+B=π−C

⇒cot(A+B)=cot(π−C)

⇒(cotA.cotB−1)/(cotA+cotB) = −cotC

⇒cotAcotB+cotBcotC+cotCcotA=1   ...(2) 

From (1) and (2) , we have,

cot²A+cot²B+cot²C+2cotAcotB+2cotBcotC+2cotCcotA

=3[cotAcotB+cotBcotC+cotCcotA]

⇒cot²A+cot²B+cot²C−cotAcotB−cotBcotC−cotCcotA = 0

⇒1/2[(cotA−cotB)²+(cotB−cotC)²+(cotC−cotA)²]=0

[∵x²+y²+z²−zx−xy−zy=1/2{(x−y)²+(y−z)²+(z−x)²}]

⇒(cotA−cotB)²+(cotB−cotC)²+(cotC−cotA)²=0 ...(3)

⇒cotA = cotB = cotC

• [In eq (3),Since  R.H.S. is zero, each square must be zero]

⇒A = B = C  [for triangle]

⇒ Triangle is equilateral. 

• [Note: a small req,pls give a ♡ ,if you find the answer helpful,it takes too much time to write these anwers..thank you _/\_ ]