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If sin⁡θ+cos⁡θ=a and cos⁡θ−sin⁡θ=b, sin⁡θ(sin⁡θ−cos⁡θ)+sin2⁡θsin2⁡θ−cos2⁡θ+sin3⁡θsin3⁡θ−cos3⁡θ+… is equal to 

a
1−ab1+ab
b
1−a23−a2
c
1−ab1+ab+1−a23−a2
d
1+ab1−ab−a2−13−a2

detailed solution

Correct option is C

We havesin2⁡θ+sin4⁡θ+sin6⁡θ+⋯.....−sin⁡θcos⁡θ+sin2⁡θcos2⁡θ+sin3⁡θ+cos3⁡θ+⋯ =sin2⁡θ1−sin2⁡θ−sin⁡θcos⁡θ1−sin⁡θcos⁡θ=1−cos⁡2θ1+cos⁡2θ−sin⁡2θ(2−sin⁡2θ)=1−ab1+ab+1−a23-a2

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