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Question

# If for 0<x<π/2, e[(sin2x+sin4x+sin6x+...+∞)loge2] satisfies the quadratic equation, x2−9x+8=0, find the value of sinx−cosxsinx+cosx

A
32
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B
2+3
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C
23
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D
3+2
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Solution

## The correct option is C 2−√3∵0<x<π/2∴0<sin2x<1Then sin2x+sin4x+sin6x+...+∞=sin2x1−sin2x=tan2x .... As series is in G.P.∴exp[(sin2x+sin4x+sin6x+...+∞)loge2]=exp[tan2xloge2]=exp{loge2tan2x}=eloge2tan2x=2tan2xLet y=2tan2x∵y satisfies quadratic equationy2−9y+8=0∴y=1,8if y=1=2tan2x∴2tan2x=20∴tan2x=0∴x=0 (impossible) ∵x>0Now if y=8=2tan2x∴2tan2x=23⇒tan2x=3∴tanx=√3∴sinx−cosxsinx+cosx=tanx−1sinx+1=√3−1√3+1×√3−1√3−1=(√3−1)23−1=3+1−2√32=2−√3

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