Prove..........☝️☝️
Attachments:
Answers
Answered by
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
Similar questions