If tan-1x−1x+2 + tan-1x+1x+2 = π4, then find the value of x.
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L.H.S. = tan-1x−1x+2 + tan-1x+1x+2
=tan−1[x−1x+2+x+1x+21−(x−1x+2)(x+1x+2)]
=tan−1[2x2−4−3]
=π4=tan−1(1)
∴2x2−4−3=1
∴x=±12
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