(cosA+cosBsinA−sinB)2008+(sinA+sinBcosA−cosB)2008=

(cosA+cosBsinA−sinB)2008+(sinA+sinBcosA−cosB)2008 [2cos(A+B2).cos(A−B2)2cos(A+B2).sin(A−B2)]2008+[2sin(A+B2).cos(A−B2)−2sin(A+B2).sin(A−B2)]2008 =cot2008(A−B2)+cot2008(A−B2) =2cot2008(A−B2)

(cosA+cosBsinA−sinB)2008+(sinA+sinBcosA−cosB)2008=

(cosA+cosBsinA−sinB)2008+(sinA+sinBcosA−cosB)2008 [2cos(A+B2).cos(A−B2)2cos(A+B2).sin(A−B2)]2008+[2sin(A+B2).cos(A−B2)−2sin(A+B2).sin(A−B2)]2008 =cot2008(A−B2)+cot2008(A−B2) =2cot2008(A−B2)

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${(\frac{\cos A + \cos B}{\sin A - \sin B})}^{2008} + {(\frac{\sin A + \sin B}{\cos A - \cos B})}^{2008} =$

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2 \tan^{2018} (\frac{A - B}{2})

2 \cot^{2008} (\frac{A - B}{2})

2 \cot^{2008} (\frac{A + B}{2})

2 \tan^{2008} (\frac{A + B}{2})

answer is B.

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Detailed Solution

{(\frac{\cos A + \cos B}{\sin A - \sin B})}^{2008} + {(\frac{\sin A + \sin B}{\cos A - \cos B})}^{2008}

${[\frac{2 \cos (\frac{A + B}{2}) . \cos (\frac{A - B}{2})}{2 \cos (\frac{A + B}{2}) . \sin (\frac{A - B}{2})}]}^{2008} + {[\frac{2 \sin (\frac{A + B}{2}) . \cos (\frac{A - B}{2})}{- 2 \sin (\frac{A + B}{2}) . \sin (\frac{A - B}{2})}]}^{2008}$