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Byju's Answer
Standard XIII
Mathematics
Evaluation of a Determinant
A = [ 1 tan...
Question
A
=
[
1
tan
x
−
tan
x
1
]
and
f
(
x
)
is defined as
f
(
x
)
=
det
.
A
T
A
−
1
then the value of
f
(
f
(
f
.
.
.
.
.
.
f
(
x
)
)
)
)
n
times
(
for
n
≥
2
)
is
Open in App
Solution
A
=
[
1
tan
x
−
tan
x
1
]
Hence,
det
A
=
sec
2
x
∴
det
A
T
=
sec
2
x
Now,
f
(
x
)
=
det
(
A
T
A
−
1
)
=
(
det
A
T
)
(
det
A
−
1
)
=
(
det
A
T
)
(
det
A
)
−
1
=
det
(
A
T
)
det
(
A
)
=
1
Hence,
f
(
x
)
=
1
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