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If
sin
2
α
−
sin
3
α
+
sin
4
α
cos
2
α
−
cos
3
α
+
cos
4
α
=
tan
k
α
\frac{\sin 2\alpha - \sin 3\alpha + \sin 4\alpha}{\cos 2\alpha - \cos 3\alpha + \cos 4\alpha} = \tan k\alpha
c
o
s
2
α
−
c
o
s
3
α
+
c
o
s
4
α
s
i
n
2
α
−
s
i
n
3
α
+
s
i
n
4
α
=
tan
k
α
…
Mathematics
Problem
If
sin
2
α
−
sin
3
α
+
sin
4
α
cos
2
α
−
cos
3
α
+
cos
4
α
=
tan
k
α
\frac{\sin 2\alpha - \sin 3\alpha + \sin 4\alpha}{\cos 2\alpha - \cos 3\alpha + \cos 4\alpha} = \tan k\alpha
c
o
s
2
α
−
c
o
s
3
α
+
c
o
s
4
α
s
i
n
2
α
−
s
i
n
3
α
+
s
i
n
4
α
=
tan
k
α
is an identity then the value of
k
k
k
is equal to - (A) 2 (B) 3 (C) 4 (D) 6
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