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Byju's Answer
Standard XII
Mathematics
Inverse of a Matrix
If A = [ 1...
Question
If
A
=
⎡
⎣
1
t
a
n
θ
2
−
t
a
n
θ
2
1
⎤
⎦
and AB = I, then B =
A
c
o
s
2
θ
2
.
A
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B
c
o
s
2
θ
2
.
A
T
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C
c
o
s
2
θ
2
.
I
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D
None of these
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Solution
The correct option is
B
c
o
s
2
θ
2
.
A
T
|
A
|
=
1
+
t
a
n
2
θ
2
=
s
e
c
2
θ
2
A
B
=
I
⇒
B
−
I
A
−
1
[
1
0
0
1
]
⎡
⎢
⎣
1
−
t
a
n
θ
2
t
a
n
θ
2
1
⎤
⎥
⎦
s
e
c
2
θ
2
=
c
o
s
2
θ
2
.
A
T
.
Suggest Corrections
12
Similar questions
Q.
If
[
1
−
t
a
n
θ
t
a
n
θ
1
]
[
1
t
a
n
θ
−
t
a
n
θ
1
]
=
[
a
−
b
−
b
a
]
, then
Q.
Show that,
⎡
⎣
1
−
tan
θ
2
tan
θ
2
1
⎤
⎦
⎡
⎣
1
tan
θ
2
−
tan
θ
2
1
⎤
⎦
−
1
=
[
cos
θ
−
sin
θ
sin
θ
cos
θ
]
.
Q.
The value of
⎡
⎢ ⎢ ⎢
⎣
1
−
tan
θ
4
tan
θ
4
1
⎤
⎥ ⎥ ⎥
⎦
⎡
⎢ ⎢ ⎢
⎣
1
tan
θ
4
−
tan
θ
4
1
⎤
⎥ ⎥ ⎥
⎦
−
1
is
Q.
If
A
(
θ
)
=
[
1
tan
θ
−
tan
θ
1
]
and
A
B
=
I
, then
(
sec
2
θ
)
B
is equal to
Q.
Prave that:
(1)
sin
2
θ
cosθ
+
cosθ
=
secθ
(2)
cos
2
θ
1
+
tan
2
θ
=
1
(3)
1
-
sinθ
1
+
sinθ
=
secθ
-
tanθ
(4)
secθ
-
cosθ
cotθ
+
tanθ
=
tanθ
secθ
(5)
cotθ
+
tanθ
=
cosecθ
secθ
(6)
1
secθ
-
tanθ
=
secθ
+
tanθ
(7)
sec
4
θ
-
cos
4
θ
=
1
-
2
cos
2
θ
(8)
secθ
+
tanθ
=
cos
θ
1
-
sinθ
(9) If
t
anθ
+
1
tanθ
=
2
, then show that
tan
2
θ
+
1
tan
2
θ
=
2
(10)
tanA
1
+
tan
2
A
2
+
cotA
1
+
cot
2
A
2
=
sin
A
cos
A
(11)
sec
4
A
1
-
sin
4
A
-
2
tan
2
A
=
1
(12)
tanθ
secθ
-
1
=
tanθ
+
secθ
+
1
tanθ
+
secθ
-
1
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