Question

a2b2+ abc -- 156c2
i)​

Answer 1

Answer:

Given:a

2

,b

2

,c

2

are in A.P

⇒b

2

−a

2

=c

2

−b

2

⇒k

2

sin

2

B−k

2

sin

2

A=k

2

sin

2

C−k

2

sin

2

B using sine rule a=ksinA,b=ksinB and c=ksinC

⇒sin(B+A)sin(B−A)=sin(C+B)sin(C−B)

⇒sin(π−C)sin(B−A)=sin(π−A)sin(C−B) since A+B+C=π

⇒sinCsin(B−A)=sinAsin(C−B)

⇒

sinA

sin(B−A)

=

sinC

sin(C−B)

⇒

sinAsinB

sin(B−A)

=

sinBsinC

sin(C−B)

sin(B−A)

=

sinC

sin(C−B)

⇒

sinAsinB

sin(B−A)

=

sinBsinC

sin(C−B)

⇒

sinAsinB

sinBcosA−cosBsinA

=

sinBsinC

sinCcosB−cosBsinC

⇒cotA−cotB=cotB−cotC

∴cotA,cotB,cotC are in A.P

hence Proved

Step-by-step explanation:

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