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Byju's Answer
Standard XII
Mathematics
Area of Triangle with Coordinates of Vertices Given
Without expan...
Question
Without expanding evaluate the determinant;
∣
∣ ∣ ∣
∣
s
i
n
α
c
o
s
α
s
i
n
(
α
+
δ
)
s
i
n
β
c
o
s
β
s
i
n
(
β
+
δ
)
s
i
n
γ
c
o
s
γ
s
i
n
(
γ
+
δ
)
∣
∣ ∣ ∣
∣
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Solution
∣
∣ ∣ ∣
∣
sin
α
cos
α
sin
(
α
+
δ
)
sin
β
cos
β
sin
(
β
+
δ
)
sin
γ
cos
γ
sin
(
γ
+
δ
)
∣
∣ ∣ ∣
∣
=
∣
∣ ∣
∣
sin
α
cos
α
sin
α
cos
δ
+
cos
α
sin
δ
sin
β
cos
β
sin
β
cos
δ
+
cos
β
+
sin
δ
sin
γ
cos
γ
sin
γ
cos
δ
+
cos
γ
sin
δ
∣
∣ ∣
∣
Using
sin
(
A
+
B
)
=
sin
A
cos
B
+
cos
A
sin
B
=
∣
∣ ∣
∣
sin
α
cos
α
sin
α
cos
δ
sin
β
cos
β
sin
β
cos
δ
sin
γ
cos
γ
sin
γ
cos
δ
∣
∣ ∣
∣
+
∣
∣ ∣
∣
sin
α
cos
α
cos
α
sin
δ
sin
β
cos
β
cos
β
sin
δ
sin
γ
cos
γ
cos
γ
sin
δ
∣
∣ ∣
∣
Using
∣
∣ ∣
∣
a
+
b
g
j
c
+
d
h
k
e
+
f
i
l
∣
∣ ∣
∣
=
∣
∣ ∣
∣
a
g
j
c
h
k
e
i
l
∣
∣ ∣
∣
+
∣
∣ ∣
∣
b
g
j
d
h
k
f
i
l
∣
∣ ∣
∣
=
cos
δ
∣
∣ ∣
∣
sin
α
cos
α
sin
α
sin
β
cos
β
sin
β
sin
γ
cos
γ
sin
γ
∣
∣ ∣
∣
+
sin
δ
∣
∣ ∣
∣
sin
α
cos
α
cos
α
sin
β
cos
β
cos
β
sin
γ
cos
γ
cos
γ
∣
∣ ∣
∣
taking
cos
δ
from third column of
1
s
t
det
sin
δ
from third column of
2
n
d
det
=
cos
δ
.0
+
sin
δ
.0
=
0
det having two same rows or columns have zero value.
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Similar questions
Q.
Evaluate the determinant
Δ
=
∣
∣ ∣
∣
1
2
4
−
1
3
0
4
1
0
∣
∣ ∣
∣
Q.
Without expanding, prove that
Δ
=
∣
∣ ∣
∣
x
+
y
y
+
z
z
+
x
z
x
y
1
1
1
∣
∣ ∣
∣
=
0
Q.
If 5
Δ
6 = 1331, 6
Δ
1 = 343, and 2
Δ
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Q.
Evaluate :
Δ
=
∣
∣ ∣
∣
cos
α
cos
β
cos
α
sin
β
−
sin
α
−
sin
β
cos
β
0
sin
α
cos
β
sin
α
sin
β
cos
α
∣
∣ ∣
∣
Q.
Evaluate :
Δ
=
∣
∣ ∣
∣
0
s
i
n
α
−
c
o
s
α
−
s
i
n
α
0
s
i
n
β
c
o
s
α
−
s
i
n
β
0
∣
∣ ∣
∣
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