Question

solve this question from trigonometry​

Answer 1

Answer:

in ∆ABC <A + <B +<C =180°

•°• A + B = 180° - <C

•°• cot (A + B ) = cot( 180° -c )

or, cotA×cotB -1 /cotB + cotA =-cotC

or , cotA × cotB -1 =-cotA×cotB ×cotC

cotA ×cotB + cotB × cotC + cotC × cotA =1

now, from question ..

cotA + cotB + cotC = √3

•°• (cotA + cotB + cotC )² =(√3)²

or , (cot²A + cot²B + cot²C +2cotA ×cotB + cotB ×cotC + cotC ×cotA ) = 3

or ,

cot²A + cot²B + cot²C +2 ×1 =3 -----1)

or, cot²A +cot²B + cot²C -1 =0

or, 2cot²A + 2cot²B + 2cot²C -2 ×1 =0

or, 2cot²A + 2cot²B + 2cot²C -2 (cotA×cotB +cotB×cotC + cotC×cotA ) =0 (from equation 1 )

or,

(cot²A + cot²B - 2cotA×cotB) + (cot²B + cot²C - 2cotB × cotC ) +(cot²A + cot²A -2 cotC × cotA)= 0

or, (cotA-cotB)² +(cotB -cotC)² + ( cotC - cotA)² =0

this only possible ,when ,

(cotA- cotB )² =0

=> cotA =cotB

and, (cotB - cotC )² =0

=> cotB = cotC

and ,

(cotC - cotA )² = 0

=> cotC =cotA

hence ,A=B =C .

therefore , ∆ABC is equilateral triangle ...