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Q.

If cos2θ+2sin2θ+3cos2θ+4sin2θ+......10 terms=30, where 0<θ<180°, then the value of cosθsinθ is

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a

0

b

1

c

312

d

1

answer is D.

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Detailed Solution

Ifcos2θ+2sin2θ+3cos2θ+4sin2θ+.....=30cos2θ(1+3+5+7+9)+sin2θ(2+4+6+8+10)=3025cos2θ+30sin2θ=3025cos2θ+25sin2θ+5sin2θ=3025+5sin2θ=305sin2θ=5sin2θ=1θ=90°[0<θ<180°]cosθsinθ=cos90°sin90°=1

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