Question 1112838
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No; it is NOT a trig identity -- so you can't prove it.<br>
The RHS as you show it is
{{{sin(b)/cos(b)-1 + sin(b)/cos(b)+1}}}
which clearly simplifies to {{{2*(sin(b)/cos(b))}}} or {{{2tan(b)}}}<br>
That is not equivalent to {{{2/sin(b)}}}<br>
What you undoubtedly meant to write on the RHS was
{{{sin(b)/(cos(b)-1) + sin(b)/(cos(b)+1)}}}
That expression can be simplified nicely; but it too is NOT equivalent to {{{2/sin(b)}}}.<br>