Question

Prove..........☝️☝️​

Answer 1

Answer:

A+BC=π (given)

⇒,C=π−(A+B)−(i)

Now,

sin  

2

A+sin  

2

B+sin  

2

C=  

2

1−cos2A

​  

+  

2

1−cos2B

​  

+  

2

1−cos2C

​  

 

[As cos2x=1−2sin  

2

x ]

=  

2

3

​  

−  

2

1

​  

[cos2A+cos2B+cos2C]

=[Using cosC+cos0=2cos  

2

C+D

​  

cos  

2

C−D

​  

]

=  

2

3

​  

−  

2

1

​  

[2cos(A+B)cos(A−B)+cos2C]

=  

2

3

​  

−  

2

1

​  

[2cos(π−C)cos(A−B)+cos2C] (From eq (i))

=  

2

3

​  

−  

2

1

​  

[−2cosCcos(A−B)+2cos  

2

C−1] (cos(π−x)=−cosx)

=  

2

3

​  

+cosCcos(A−B)−cos  

2

C+  

2

1

​  

 [multiplying  

2

1

​  

 inside bracket]

=2+cosC[cos(A−B)−cosC]

=2+cosC[cos(A−B)−cos(π−(A+B))] [From eq (i)]

=2+cosC[cos(A−B)+cos(A+B)]

=2+cosC×[2cosAcosB]

=2+2cosAcosBcosC