sin(n+1)A-sin(n-1)A/cos(n+1)A+2cosnA+cos(n-1)A
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hence answer is tanA
sin(n+1)A-sin(n-1)A/cos(n+1)A+2cosA+cos(n-1)A
= sin nA +sinA-sin nA+sinA/cosnA+cosA+2cosA+cosnnA-cosA
=2sinA/2cosA
=tanA
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