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Question
Mathematics
If [1&- tanθ tan θ&1][1& tan θ - tan θ &1]-1 = [a&-b b&a] then
Q. If
[
1
tan
θ
​
−
tan
θ
1
​
]
[
1
−
tan
θ
​
tan
θ
1
​
]
−
1
=
[
a
b
​
−
b
a
​
]
then
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A
a
=
1
=
b
22%
B
a
=
cos
2
θ
,
b
=
sin
2
θ
50%
C
a
=
sin
2
θ
,
b
=
cos
2
θ
23%
D
a
=
cos
θ
,
b
=
sin
θ
4%
Solution:
[
1
tan
θ
​
−
tan
θ
1
​
]
[
1
−
tan
θ
​
tan
θ
1
​
]
−
1
=
[
a
b
​
−
b
a
​
]
Since
A
−
1
=
∣
A
∣
a
d
j
A
​
∴
Â
[
1
tan
θ
​
−
tan
θ
1
​
]
1
+
t
a
n
2
θ
[
1
tan
θ
​
−
tan
θ
1
​
]
​
=
[
a
b
​
−
b
a
​
]
1
+
t
a
n
2
θ
1
​
[
1
−
tan
2
θ
tan
θ
+
tan
θ
​
−
tan
θ
−
tan
θ
−
tan
2
θ
+
1
​
]
=
[
a
b
​
−
b
a
​
]
[
1
+
t
a
n
2
θ
1
−
t
a
n
2
θ
​
1
+
t
a
n
2
θ
2
t
a
n
θ
​
​
1
+
t
a
n
2
θ
−
2
t
a
n
θ
​
1
+
t
a
n
2
θ
1
−
t
a
n
2
θ
​
​
]
=
[
a
b
​
−
b
a
​
]
⇒
[
cos
2
θ
sin
2
θ
​
−
sin
2
θ
cos
2
θ
​
]
=
[
a
b
​
−
b
a
​
]
So,
a
=
cos
2
θ
,
b
=
sin
2
θ