plz prove this..........
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Answer:
PLZ MARK MY ANSWER AS BRAINLIEST.
Step-by-step explanation:
The first part of
L
H
S
=
a
3
sin
(
B
−
C
)
=
(
2
R
sin
A
)
3
sin
(
B
−
C
)
=
2
R
3
⋅
2
sin
2
A
⋅
2
sin
A
⋅
sin
(
B
−
C
)
=
2
R
3
⋅
(
1
−
cos
2
A
)
⋅
[
cos
(
A
−
B
+
C
)
−
cos
(
A
+
B
−
C
)
]
=
2
R
3
⋅
(
1
−
cos
2
A
)
⋅
[
cos
(
π
−
2
B
)
−
cos
(
π
−
2
C
)
]
=
2
R
3
⋅
(
1
−
cos
2
A
)
⋅
(
cos
2
C
−
cos
2
B
)
=
2
R
3
⋅
(
cos
2
C
−
cos
2
B
−
cos
2
A
⋅
cos
2
C
+
cos
2
A
⋅
cos
2
B
)
Similarly, the second part
=
2
R
3
(
cos
2
A
−
cos
2
C
−
cos
2
A
⋅
cos
2
B
+
cos
2
B
⋅
cos
2
C
)
And the third part
=
2
R
3
(
cos
2
B
−
cos
2
A
−
cos
2
B
⋅
cos
2
C
+
cos
2
A
⋅
cos
2
C
)
Adding up all these three parts, we get,
a
3
sin
(
B
−
C
)
+
b
3
sin
(
C
−
A
)
+
c
3
sin
(
A
−
B
)
=
0
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