Q.

The general solution of the equation 1−sinx+....+−1nsinnx+....1+sinx+....+sinnx+.....=1−cos2x1+cos2x ,  x≠2n+1π/2,n∈I is

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a

−1nπ/3+nπ

b

−1nπ/6+nπ

c

−1n+1π/6+nπ

d

−1n−1π/3+nπ

answer is B.

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Detailed Solution

We have 1−sinx+....+−1nsinnx+....1+sinx+....+sinnx+.....=1−cos2x1+cos2xNumerator and denominator both are in GP S∞=a1-r⇒11+sinx×1−sinx1=2sin2x2cos2x,∵−1
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