Question

In any ΔABC, if a². b², c² are are in A.P.. then prove that cot A, cot B, cot C are in A.P.

Answer 1

hope both the methods are helpful to you .. :))

Answer 2

Answer:

Proved

Step-by-step explanation:

In any ΔABC, if a². b², c² are are in A.P.. then prove that cot A, cot B, cot C are in A.P.

a². b², c² are are in A.P

=> 2b² = a² + c²  - eq 1

in ΔABC

a² = b² + c² - 2bcCosA

b² = a² + c² - 2acCosB

c² = a² + b² - 2abCosC

putting these values in eq 1

=> 2(a² + c² - 2acCosB) = (b² + c² - 2bcCosA) + (a² + b² - 2abCosC)

=> 2a² + 2c² - 4acCosB = 2b² + a² + c² - 2bcCosA - 2abCosC

=> a² + c² - 2b²   - 4acCosB =  - 2bcCosA - 2abCosC

2b² = a² + c² => a² + c² - 2b² = 0

=>  - 4acCosB =  - 2bcCosA - 2abCosC  

using CotxSinx = Cosx  & dividing by - 2

=> 2acCotBSinB = bcCotASinA  + abCotCSinC   - eq 2

in  ΔABC

a/SinA = b/SinB = c/SinC  = K

=> SinA = a/K   , SinB = b/K   SinC = c/K

putting in eq 2

2acCotBb/K  = bcCotAa/K  + abCotCc/K

=> (abc/K)2CotB = (abc/K)CotA + (abc/k)CotC

=>(abc/K)2CotB = (abc/k) (CotA + cotC)

Cancelling abc/k from both sides

=> 2CotB = CotA + CotB

Hence cot A, cot B, cot C are in A.P.

QED