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Question

# If αϵ[π2,π] then the value of √1+sinα−√1−sinα is equal to

A
2cosα2
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B
2sinα2
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C
2
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D
none of these
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Solution

## The correct option is A 2cosα2Given α∈[π2,π]⟹α2∈[π4,π2]⟹−α2∈[−π2,−π4]⟹π4−α2∈[−π4,0]⟹cos(π4−α2)>0 and sin(π4−α2)<0√1+sinα−√1−sinα⟹√1+cos(π2−α)−√1−cos(π2−α) (sinθ=cos(π2−cosθ)) ⟹√2cos2(π4−α2)−√2sin2(π4−α2) (1+cosθ=2cos2θ,1−cosθ=2sin2θ)⟹√2∣∣∣cos(π4−α2)∣∣∣−√2∣∣∣sin(π4−α2)∣∣∣⟹√2(cos(π4−α2)+sin(π4−α2)) (|x|={xx≥0−xx>0)⟹√2(cosπ4cosα2+sinπ2sinα2+sinπ4cosα2−cosπ4sinα2) (cos(A−B)=cosAcosB+sinAsinB,sin(A−B)=sinAcosB−cosAsinB)⟹√2⎛⎜ ⎜⎝cosα2+sinα2√2+cosα2−sinα2√2⎞⎟ ⎟⎠⟹cosα2+sinα2+cosα2−sinα2⟹2cosα2

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