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Problem
Problem
If
tan
(
3
π
11
)
+
4
sin
(
2
π
11
)
=
λ
\tan\left(\frac{3\pi}{11}\right) + 4 \sin\left(\frac{2\pi}{11}\right) = \lambda
tan
(
11
3
π
)
+
4
sin
(
11
2
π
)
=
λ
, then the value of
1
+
λ
2
+
λ
4
+
λ
6
1 + \lambda^2 + \lambda^4 + \lambda^6
1
+
λ
2
+
λ
4
+
λ
6
must be.
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