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Dear Student
If α,β0,π2Since 0<α<π20<α2<π40<π4-α2<π40<tanπ4-α2<1-----(1)Since 0<β<π20<β2<π40<π4-β2<π40<tanπ4-β2<1-----(2)From eq(1) and eq(2) we get0<tanπ4-α2tanπ4-β2<1Since 2tan-1x=2x1-x2,where x(0,1)Given LHS of the equation2tan-1tanπ4-α2tanπ4-β2=tan-12tanπ4-α2tanπ4-β21-tan2π4-α2tan2π4-β2We know that tan(A-B)=tanA-tanB1+tanAtanBso the equation becomes=tan-121-tanα21+tanα21-tanβ21+tanβ21-1-tanα21+tanα221-tanβ21+tanβ22=tan-121-tanα21-tanβ21+tanα21+tanβ21+tanα221+tanβ22-1-tanα221-tanβ221+tanα221+tanβ22=tan-121-tanα21-tanβ21+tanα21+tanβ21+tan2α2+2tanα21+tan2β2+2tanβ2-1+tan2α2-2tanα21+tan2β2-2tanβ2=tan-121-tan2α21-tan2β21+tan2β2+2tanβ2+tan2α2+tan2α2tan2β2+2tan2α2tanβ2+2tanα2+2tanα2tan2β2+4tanα2tanβ2-1+tan2β2-2tanβ2+tan2α2+tan2α2tan2β2-2tan2α2tanβ2-2tanα2-2tanα2tan2β2+4tanα2tanβ2By simplifying we get=tan-121-tan2α21-tan2β24tanβ2+4tan2α2tanβ2+4tanα2+4tanα2tan2β2=tan-121-tan2α21-tan2β24tanβ21+tan2α2+tanα21+tan2β2Now divinding  both numerator and denominator with 1+tan2α21+tan2β2 we get=tan-11-tan2α21-tan2β21+tan2α21+tan2β22tanβ21+tan2α2+tanα21+tan2β21+tan2α21+tan2β2=tan-11-tan2α21+tan2α21-tan2β21+tan2β22tanβ21+tan2β2+2tanα21+tan2α2we know thatcos2θ=1-tan2θ1+tan2θand sin2θ=2tanθ1+tan2θso we get=tan-1cosαcosβsinβ+sinα=RHS
                                                                   Hence proved
  Regards

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