Trigonometric Substitution Maybe?

$\\ \frac { 1 }{ x } +\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } =\frac { 35 }{ 12 } ......(1)\\ Put,x=\frac { 1 }{ u } ,\sqrt { 1-{ x }^{ 2 } } =\frac { 1 }{ v } .......(2)\\ from\quad (1)\quad and\quad (2)\quad we\quad get,\\ u+v=\frac { 35 }{ 12 } ,{ u }^{ 2 }+{ v }^{ 2 }={ u }^{ 2 }{ v }^{ 2 }\\ \Rightarrow uv=\frac { 25 }{ 12 } ,\frac { -49 }{ 12 } \\ When\quad uv=\frac { 25 }{ 12 } ,\quad we\quad get\quad x=\frac { 3 }{ 5 } ,\frac { 4 }{ 5 } \\ When\quad uv=\frac { -49 }{ 12 } ,\quad we\quad get\quad x=\frac { -(5+\sqrt { 73 } ) }{ 14 } \\ \therefore \quad Sum\quad of\quad real\quad values\quad of\quad x=\frac { 3 }{ 5 } +\frac { 4 }{ 5 } +\frac { -(5+\sqrt { 73 } ) }{ 14 } =\boxed { 0.43257 }$

$Let x=CosA, \ so \ SecA+CosecA=\dfrac{35}{12},\ so\ Cos^{-1}\frac 3 5,\ Cos^{-1}\frac 4 5.\\ \implies\ x=\frac 3 5, \frac 4 5.\\ Manipulating \ and\ Simplifying,\ we\ get\ a\ fourth\ degree\ equation.\\ \text{Dividing this equation by (x-.6)(x-.8) and solving we get x= -0.96763}.\\ \therefore \ the\ sum =0.43237.\\$
The values can be checked on a graph.