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Q.

If m=Sinθ+Cosθ; n=Secθ+Cosecθ then

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a

nm2-1=2m

b

nm2+1=2m

c

2nm2-1=m

d

2nm2+1=m

answer is B.

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Detailed Solution

m=sinθ+cosθm2=1+2sinθ cosθsinθ cosθ=m2-12n=secθ+cosecθn2=sec2θ+cosec2θ+2secθ cosecθn2=1sin2θ cos2θ+2sinθ cosθn2=m2m2-122n=2mm2-1nm2-1=2m

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