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Byju's Answer
Standard XII
Mathematics
Derivative of One Function w.r.t Another
Without expan...
Question
Without expanding, show that the values of each of the following determinants are zero:
(i)
8
2
7
12
3
5
16
4
3
(ii)
6
-
3
2
2
-
1
2
-
10
5
2
(iii)
2
3
7
13
17
5
15
20
12
(iv)
1
/
a
a
2
b
c
1
/
b
b
2
a
c
1
/
c
c
2
a
b
(v)
a
+
b
2
a
+
b
3
a
+
b
2
a
+
b
3
a
+
b
4
a
+
b
4
a
+
b
5
a
+
b
6
a
+
b
(vi)
1
a
a
2
-
b
c
1
b
b
2
-
a
c
1
c
c
2
-
a
b
(vii)
49
1
6
39
7
4
26
2
3
(viii)
0
x
y
-
x
0
z
-
y
-
z
0
(ix)
1
43
6
7
35
4
3
17
2
(x)
1
2
2
2
3
2
4
2
2
2
3
2
4
2
5
2
3
2
4
2
5
2
6
2
4
2
5
2
6
2
7
2
(xi)
a
b
c
a
+
2
x
b
+
2
y
c
+
2
z
x
y
z
(xii)
2
x
+
2
-
x
2
2
x
-
2
-
x
2
1
3
x
+
3
-
x
2
3
x
-
3
-
x
2
1
4
x
+
4
-
x
2
4
x
-
4
-
x
2
1
(xiii)
sin
α
cos
α
cos
(
α
+
δ
)
sin
β
cos
β
cos
(
β
+
δ
)
sin
γ
cos
γ
cos
(
γ
+
δ
)
(xiv)
sin
2
23
°
sin
2
67
°
cos
180
°
-
sin
2
67
°
-
sin
2
23
°
cos
2
180
°
cos
180
°
sin
2
23
°
sin
2
67
°
(xv)
cos
x
+
y
-
sin
x
+
y
cos
2
y
sin
x
cos
x
sin
y
-
cos
x
sin
x
-
cos
y
(xvi)
23
+
3
5
5
15
+
46
5
10
3
+
115
15
5
(xvii)
sin
2
A
cot
A
1
sin
2
B
cot
B
1
sin
2
C
cot
C
1
,
where
A
,
B
,
C
are
the
angles
of
∆
A
B
C
.
Open in App
Solution
(
i
)
∆
=
8
2
7
12
3
5
16
4
3
=
0
2
7
0
3
5
0
4
3
Applying
C
1
→
C
1
-
4
C
2
⇒
∆
=
0
(
i
i
)
∆
=
6
-
3
2
2
-
1
2
-
10
5
2
=
0
-
3
2
0
-
1
2
0
5
2
Applying
C
1
→
C
1
+
2
C
2
⇒
∆
=
0
(
i
i
i
)
∆
=
2
3
7
13
17
5
15
20
12
=
2
3
7
13
17
5
13
17
5
Applying
R
3
→
R
3
-
R
1
⇒
∆
=
0
(
i
v
)
∆
=
1
a
a
2
b
c
1
b
b
2
a
c
1
c
c
2
a
b
=
1
a
3
a
b
c
1
b
3
a
b
c
1
c
3
a
b
c
Applying
R
1
→
a
R
1
,
R
2
→
b
R
2
and
R
3
→
c
R
3
=
a
b
c
1
a
3
1
1
b
3
1
1
c
3
1
⇒
∆
=
0
(
v
)
∆
=
a
+
b
2
a
+
b
3
a
+
b
2
a
+
b
3
a
+
b
4
a
+
b
4
a
+
b
5
a
+
b
6
a
+
b
=
a
a
a
2
a
2
a
2
a
4
a
+
b
5
a
+
b
6
a
+
b
Applying
R
1
→
R
2
-
R
1
and
R
2
→
R
3
-
R
2
=
2
a
a
a
a
a
a
4
a
+
b
5
a
+
b
6
a
+
b
=
0
(
v
i
)
∆
=
1
a
a
2
-
b
c
1
b
b
2
-
a
c
1
c
c
2
-
a
b
=
0
a
-
b
a
2
-
b
c
-
b
2
+
a
c
0
b
-
c
b
2
-
a
c
-
c
2
+
a
b
1
c
c
2
-
a
b
Applying
R
1
→
R
1
-
R
2
,
R
2
→
R
2
-
R
3
=
0
a
-
b
a
-
b
a
+
b
+
c
a
-
b
0
b
-
c
b
-
c
b
+
c
+
a
b
-
c
1
c
c
2
-
a
b
=
a
-
b
b
-
c
0
1
a
+
b
+
c
0
1
a
+
b
+
c
1
c
c
2
-
a
b
⇒
∆
=
0
(
v
i
i
)
∆
=
49
1
6
39
7
4
26
2
3
=
1
1
6
7
7
4
2
2
3
Applying
C
1
→
C
1
-
8
C
3
⇒
∆
=
0
(
v
i
i
i
)
∆
=
0
x
y
-
x
0
z
-
y
-
z
0
=
x
y
z
x
y
z
0
x
y
-
x
0
z
-
y
-
z
0
=
1
x
y
z
0
x
z
y
z
-
x
y
0
z
y
-
y
x
-
z
x
0
=
1
x
y
z
-
2
x
y
0
2
y
z
-
x
y
0
z
y
-
y
x
-
z
x
0
Applying
R
1
→
R
1
+
R
2
+
R
3
=
1
x
y
z
0
0
0
-
x
y
0
z
y
-
y
x
-
z
x
0
=
0
Applying
R
1
→
R
1
-
2
R
2
(
i
x
)
∆
=
1
43
6
7
35
4
3
17
2
=
1
1
6
7
7
4
3
3
2
=
0
Applying
C
2
→
C
2
-
7
C
3
x
)
∆
=
1
2
2
2
3
2
4
2
2
2
3
2
4
2
5
2
3
2
4
2
5
2
6
2
4
2
5
2
6
2
7
2
=
1
4
9
16
4
9
16
25
9
16
25
36
16
25
36
49
=
1
4
9
16
4
9
16
25
5
7
9
11
7
9
11
13
Applying
R
3
→
R
3
-
R
2
and
R
4
→
R
4
-
R
3
=
1
4
9
16
4
9
16
25
7
9
11
13
7
9
11
13
=
0
Applying
R
3
→
2
+
R
3
xi
)
∆
=
a
b
c
a
+
2
x
b
+
2
y
c
+
2
z
x
y
z
=
a
+
2
x
b
+
2
y
c
+
2
z
a
+
2
x
b
+
2
y
c
+
2
z
x
y
z
Applying
R
1
→
R
1
+
2
R
3
=
0
0
0
a
+
2
x
b
+
2
y
c
+
2
z
x
y
z
=
0
Applying
R
1
→
R
1
-
R
2
(xii)
2
x
+
2
-
x
2
2
x
-
2
-
x
2
1
3
x
+
3
-
x
2
3
x
-
3
-
x
2
1
4
x
+
4
-
x
2
4
x
-
4
-
x
2
1
=
2
2
x
+
2
-
2
x
+
2
2
2
x
+
2
-
2
x
-
2
1
3
2
x
+
3
-
2
x
+
2
3
2
x
+
3
-
2
x
-
2
1
4
2
x
+
4
-
2
x
+
2
4
2
x
+
4
-
2
x
-
2
1
=
4
2
2
x
+
2
-
2
x
-
2
1
4
3
2
x
+
3
-
2
x
-
2
1
4
4
2
x
+
4
-
2
x
-
2
1
Applying
C
1
→
C
1
-
C
2
=
4
1
2
2
x
+
2
-
2
x
-
2
1
1
3
2
x
+
3
-
2
x
-
2
1
1
4
2
x
+
4
-
2
x
-
2
1
=
0
(xiii)
sin
α
cos
α
cos
(
α
+
δ
)
sin
β
cos
β
cos
(
β
+
δ
)
sin
γ
cos
γ
cos
(
γ
+
δ
)
=
sin
α
sin
δ
cos
α
cos
δ
cos
(
α
+
δ
)
sin
β
sin
δ
cos
β
cos
δ
cos
(
β
+
δ
)
sin
γ
sin
δ
cos
γ
cos
δ
cos
(
γ
+
δ
)
Applying
C
1
→
sin
δ
C
1
and
C
2
→
cos
δ
C
2
=
sin
α
sin
δ
cos
(
α
+
δ
)
cos
(
α
+
δ
)
sin
β
sin
δ
cos
(
β
+
δ
)
cos
(
β
+
δ
)
sin
γ
sin
δ
cos
(
γ
+
δ
)
cos
(
γ
+
δ
)
Applying
C
2
→
C
2
-
C
1
=
0
(xiv)
sin
2
23
°
sin
2
67
°
cos
180
°
-
sin
2
67
°
-
sin
2
23
°
cos
2
180
°
cos
180
°
sin
2
23
°
sin
2
67
°
=
sin
2
23
°
sin
2
90
-
23
°
-
1
-
sin
2
90
-
23
°
-
sin
2
23
°
1
-
1
sin
2
23
°
sin
2
90
-
23
°
=
sin
2
23
°
cos
2
23
°
-
1
-
cos
2
23
°
-
sin
2
23
°
1
-
1
sin
2
23
°
cos
2
23
°
=
sin
2
23
°
+
cos
2
23
°
cos
2
23
°
-
1
-
cos
2
23
°
-
sin
2
23
°
-
sin
2
23
°
1
-
1
+
sin
2
23
°
sin
2
23
°
cos
2
23
°
A
p
p
l
y
i
n
g
C
1
→
C
1
+
C
2
=
1
1
-
1
-
1
-
sin
2
23
°
1
-
cos
2
23
°
sin
2
23
°
cos
2
23
°
=
-
1
-
1
1
-
1
1
-
sin
2
23
°
1
cos
2
23
°
sin
2
23
°
cos
2
23
°
=
0
(xv)
cos
x
+
y
-
sin
x
+
y
cos
2
y
sin
x
cos
x
sin
y
-
cos
x
sin
x
-
cos
y
=
1
sin
y
cos
y
cos
x
+
y
-
sin
x
+
y
cos
2
y
sin
x
sin
y
cos
x
sin
y
sin
2
y
-
cos
x
cos
y
sin
x
cos
y
-
cos
2
y
Applying
R
2
→
sin
y
R
2
and
R
3
→
cos
y
R
3
=
1
sin
y
cos
y
cos
x
+
y
-
sin
x
+
y
cos
2
y
sin
x
sin
y
-
cos
x
cos
y
cos
x
sin
y
+
sin
x
cos
y
sin
2
y
-
cos
2
y
-
cos
x
cos
y
sin
x
cos
y
-
cos
2
y
Applying
R
2
→
R
2
+
R
3
=
-
1
sin
y
cos
y
cos
x
+
y
-
sin
x
+
y
cos
2
y
cos
x
+
y
-
sin
x
+
y
cos
2
y
-
cos
x
cos
y
sin
x
cos
y
-
cos
2
y
=
0
(xvi)
23
+
3
5
5
15
+
46
5
10
3
+
115
15
5
=
3
5
5
15
5
10
3
15
5
+
23
5
5
46
5
10
115
15
5
=
3
1
5
5
5
5
10
3
15
5
+
23
1
5
5
2
5
10
5
15
5
=
3
×
5
1
1
5
5
5
10
3
3
5
+
23
×
5
1
5
1
2
5
2
5
15
5
=
0
+
0
=
0
(xvii)
sin
2
A
cot
A
1
sin
2
B
cot
B
1
sin
2
C
cot
C
1
=
sin
2
A
-
sin
2
B
cot
A
-
cot
B
0
sin
2
B
cot
B
1
sin
2
C
-
sin
2
B
cot
C
-
cot
B
0
Applying
R
1
→
R
1
-
R
2
and
R
3
→
R
3
-
R
2
=
sin
A
+
B
sin
A
-
B
cos
A
sin
B
-
cos
B
sin
A
sin
A
sin
B
0
sin
2
B
cot
B
1
sin
C
+
B
sin
C
-
B
cos
C
sin
B
-
cos
B
sin
C
sin
B
sin
C
0
=
sin
π
-
C
sin
A
-
B
-
sin
A
-
B
sin
A
sin
B
0
sin
2
B
c
o
t
B
1
sin
π
-
A
sin
C
-
B
-
sin
C
-
B
sin
B
sin
C
0
∵
A
+
B
+
C
=
π
=
sin
C
sin
A
-
B
-
sin
A
-
B
sin
A
sin
B
0
sin
2
B
cos
B
sin
B
1
sin
A
sin
C
-
B
-
sin
C
-
B
sin
B
sin
C
0
=
sin
A
-
B
sin
C
-
B
sin
B
sin
C
-
1
sin
A
0
sin
2
B
cos
B
1
sin
A
-
1
sin
C
0
=
sin
A
-
B
sin
C
-
B
sin
B
sin
A
sin
C
sin
C
sin
A
-
1
0
sin
2
B
cos
B
1
sin
A
sin
C
-
1
0
Applying
R
1
→
sin
A
R
1
and
R
3
→
sin
C
R
3
=
sin
A
-
B
sin
C
-
B
sin
B
sin
A
sin
C
0
0
0
sin
2
B
cos
B
1
sin
A
sin
C
-
1
0
Applying
R
1
→
R
1
-
R
3
=
0
Suggest Corrections
0
Similar questions
Q.
Without expanding, show that the values of each of the following determinants are zero:
(i)
8
2
7
12
3
5
16
4
3
(ii)
6
-
3
2
2
-
1
2
-
10
5
2
(iii)
2
3
7
13
17
5
15
20
12
(iv)
1
/
a
a
2
b
c
1
/
b
b
2
a
c
1
/
c
c
2
a
b
(v)
a
+
b
2
a
+
b
3
a
+
b
2
a
+
b
3
a
+
b
4
a
+
b
4
a
+
b
5
a
+
b
6
a
+
b
(vi)
1
a
a
2
-
b
c
1
b
b
2
-
a
c
1
c
c
2
-
a
b
(vii)
49
1
6
39
7
4
26
2
3
(viii)
0
x
y
-
x
0
z
-
y
-
z
0
(ix)
1
43
6
7
35
4
3
17
2
(x)
1
2
2
2
3
2
4
2
2
2
3
2
4
2
5
2
3
2
4
2
5
2
6
2
4
2
5
2
6
2
7
2
(xi)
a
b
c
a
+
2
x
b
+
2
y
c
+
2
z
x
y
z
Q.
Without expanding , find the value of
∣
∣ ∣
∣
a
+
b
2
a
+
b
3
a
+
b
2
a
+
b
3
a
+
b
4
a
+
b
4
a
+
b
5
a
+
b
6
a
+
b
∣
∣ ∣
∣
Q.
If
x
,
y
,
z
∈
R
then find Determinant
⎛
⎜ ⎜ ⎜
⎝
(
2
x
+
2
−
x
)
2
(
2
x
−
2
−
x
)
2
1
(
3
x
+
3
−
x
)
2
(
3
x
−
3
−
x
)
2
1
(
4
x
+
4
−
x
)
2
(
4
x
−
4
−
x
)
2
1
⎞
⎟ ⎟ ⎟
⎠
Q.
sin
2
23
°
+
sin
2
67
°
cos
2
13
°
+
cos
2
77
°
+
sin
2
59
°
+
cos
59
°
sin
21
°
=
?
(a) 3
(b) 2
(c) 1
(d) 0
Q.
If x, y, z ∈ R, the value of the determinant
2
x
+
2
-
x
2
2
x
-
2
-
x
2
1
3
x
+
3
-
x
2
3
x
-
3
-
x
2
1
4
x
+
4
-
x
2
4
x
-
4
-
x
2
1
is equal to ________________.
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