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Problems
Contests
National and Regional Contests
USA Contests
MAA AMC
AMC 12/AHSME
1991 AMC 12/AHSME
21
21
Part of
1991 AMC 12/AHSME
Problems
(1)
Trig identities
Source: AHSME 1991 problem 21
10/30/2011
If
f
(
x
x
−
1
)
=
1
x
f\left(\frac{x}{x - 1}\right) = \frac{1}{x}
f
(
x
−
1
x
)
=
x
1
for all
x
≠
0
,
1
x \ne 0,1
x
=
0
,
1
and
0
<
θ
<
π
2
0 < \theta < \frac{\pi}{2}
0
<
θ
<
2
π
, then
f
(
sec
2
θ
)
=
f(\sec^{2}\theta) =
f
(
sec
2
θ
)
=
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
sin
2
θ
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
cos
2
θ
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
tan
2
θ
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
cot
2
θ
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
csc
2
θ
<span class='latex-bold'>(A) </span>\sin^{2}\theta\qquad<span class='latex-bold'>(B) </span>\cos^{2}\theta\qquad<span class='latex-bold'>(C) </span>\tan^{2}\theta\qquad<span class='latex-bold'>(D) </span>\cot^{2}\theta\qquad<span class='latex-bold'>(E) </span>\csc^{2}\theta
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
sin
2
θ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
cos
2
θ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
tan
2
θ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
cot
2
θ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
csc
2
θ
trigonometry
AMC