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Problem
Problem
α
2
1
−
tan
2
x
=
sin
2
x
+
α
2
−
2
cos
2
x
\frac{\alpha^2}{1 - \tan^2 x} = \frac{\sin^2 x + \alpha^2 - 2}{\cos 2x}
1
−
t
a
n
2
x
α
2
=
c
o
s
2
x
s
i
n
2
x
+
α
2
−
2
has a solution if a.
α
≤
−
1
\alpha \le -1
α
≤
−
1
b.
α
≥
1
\alpha \ge 1
α
≥
1
c.
α
=
1
/
2
\alpha = 1/2
α
=
1/2
d.
α
\alpha
α
is any real number
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