# "We want to solve:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad tan^{-1} (x) \ = \ cos^{-1} (x). #
# "We can proceed as follows:" #
# \qquad \qquad \qquad \qquad \qquad \quad tan[ tan^{-1} (x) ] \ = \ tan[ cos^{-1} (x) ]. \qquad \qquad \qquad \qquad \qquad \ (I) #
# \qquad \quad \ rArr \quad "now using:" \qquad tan^{-1}( tan ( x ) ) = x, \quad "in" \ \ (I) \quad rArr #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x \ = \ tan[ cos^{-1} (x) ]. #
# "Emphasizing:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x \ = \ tan[ \underbrace{ cos^{-1} (x) }_{\theta} ] \qquad \qquad \qquad \qquad \qquad \ (II) #
# \qquad \quad \ rArr \quad "now using:" \qquad 1 + tan^2 \theta = sec^2 \theta \quad rArr #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ tan \theta = sqrt{ sec^2 \theta - 1 } \quad rArr #
#\qquad \qquad \qquad \quad \ \ \ tan \theta = sqrt{ 1/{ cos^2 \theta } - 1 }, \quad "in" \ \ (II) \quad rArr #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad x \ = \ sqrt{ 1/{ cos^2 ( cos^{-1} (x) ) } - 1 } #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad x \ = \ sqrt{ 1/{ [ cos ( cos^{-1} (x) ) ]^2 - 1 } }. \qquad \qquad \qquad \qquad \quad (III) #
# \qquad \quad \ rArr \quad "now using:" \qquad cos^{-1}( cos( x ) ) = x, \quad "in" \ \ (III) \quad rArr #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x \ = \ sqrt{ 1/{ x^2 ) - 1 }. #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \quad x^2 \ = \ 1/{ x^2 ) - 1. #
# \qquad :. \qquad \qquad \qquad \qquad \ \ x^2 [ x^2 ] \ = \ x^2 [ 1/{ x^2 ) - 1 ]. #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x^4 \ = \ 1 - x^2. #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad x^4 + x^2 - 1 \ = \ 0. #
# :. \qquad \qquad \qquad \qquad \qquad \qquad \quad ( x^2 )^2 + ( x^2 ) - 1 \ = \ 0. \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (IV) #
# "Now let (for convenience): " \qquad u \ = \ x^2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (V) #
# "So (IV) becomes: " #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad u^2 + u - 1 \ = \ 0. #
# "So by the Quadratic Formula: " #
# \qquad \qquad \qquad \qquad \qquad \qquad \quad u \ = \ { -1 \pm sqrt{ (1)^2 - 4 (1) (-1) } }/ (2 cdot 1 ). #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad \ u \ = \ { -1 \pm sqrt{ 1 + 4 } }/2. #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad \quad \ u \ = \ { -1 \pm sqrt{ 5 } }/2. #
# \qquad :. \qquad u \ = \ { -1 - sqrt{ 5 } }/2 \qquad \qquad "or \qquad \qquad u \ = \ { -1 + sqrt{ 5 } }/2. #
# "So by (V): " #
# \qquad \qquad \qquad x^2 \ = \ - { 1 + sqrt{ 5 } }/2 \qquad \qquad "and" \qquad \qquad x^2 \ = \ { -1 + sqrt{ 5 } }/2. #
# "Looking at the original equation, at the top, we see" \ x \ "must be" #
# "a real number because the domain of" \ \ tan(x) \ \ "is, in particular," #
# "a subset of the real numbers. The first pair of equations in the" #
# "previous yields no real solutions for" \ x, \ "as" \ - { 1 + sqrt{ 5 } }/2 \ "is" #
# "negative. And the second pair yields two real solutions, as" #
# \ { -1 + sqrt{ 5 } }/2 \ "is positive [ certainly" \ sqrt{ 5 } > 1 \ "]." #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad x^2 \ = \ { -1 + sqrt{ 5 } }/2, \qquad "only". #
# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad x \ = \ \pm sqrt{ { -1 + sqrt{ 5 } }/2 }. #
# \qquad :. \qquad \qquad \qquad \qquad \quad x \ = \ \pm sqrt{ { 2 ( -1 + sqrt{ 5 } ) }/{ 2 cdot 2 } }. #
# \qquad :. \qquad \qquad \qquad \qquad \qquad x \ = \ \pm sqrt{ -2 + 2sqrt{ 5 } }/{ 2 }. #
# "These are our solutions !!" #
# "So, summarizing:"#
# \qquad \quad o. \ "the solutions of:" \qquad tan^{-1} (x) = cos^{-1} (x) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ "are:" \qquad \qquad \quad \ \ x \ = \ \pm sqrt{ -2 + 2sqrt{ 5 } }/{ 2 } \quad. \qquad \qquad square #
