Find the value of tan12[sin−12x1+x2+cos−11−y21+y2] |x|<1,y>0andxy<1
A
x+y1−xy
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B
π+x+y1−xy
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C
1−xyx+y
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D
π+1−xyx+y
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Solution
The correct option is Ax+y1−xy Let x=tanθ, then θ=tan−1x. ∴sin−12x1+x2=sin−1(2tanθ1+tan2θ) =sin−1(sin2θ)=2θ=2tan−1x Let y=tanϕ,Thenϕ=tan−1y ∴cos−11−y21+y2=cos−1(1−tan2ϕ1+tan2ϕ) =cos−1(cos2ϕ)=2ϕ=2tan−1y ∴tan12[sin−12x1+x2+cos−11−y21+y2] =tan12[2tan−1x+2tan−1y] =tan[tan−1x+tan−1y] =tan[tan−1(x+y1−xy)]=x+y1−xy