Solve for $$x:\,\sqrt {x + 1} - \sqrt {x - 1} = 1.$$

Solve for $$x:{4^x} - {3^{^{x - {1 \over 2}}}}\, = {3^{^{x + {1 \over 2}}}}\, - {2^{2x - 1}}$$

If $$\left( {m\,,\,n} \right) = {{\left( {1 - {x^m}} \right)\left( {1 - {x^{m - 1}}} \right).......\left( {1 - {x^{m - n + 1}}} \right)} \over {\left( {1 - x} \right)\left( {1 - {x^2}} \right).........\left( {1 - {x^n}} \right)}}$$

where $$m$$ and $$n$$ are positive integers $$\left( {n \le m} \right),$$ show that
$$\left( {m,n + 1} \right) = \left( {m - 1,\,n + 1} \right) + {x^{m - n - 1}}\left( {m - 1,n} \right).$$

If $$\tan \alpha = {m \over {m + 1}}\,$$ and $$\tan \beta = {2 \over {2m + 1}},$$ find the possible values of $$\left( {\alpha + \beta } \right).$$