Show taht `2tan^-1(tan(alpha/2)tan(pi/4-beta/2))=tan^-1((sinalphacosbeta)/(cosalpha+sinbeta))`

Explanation

We know that,

` 2 tan ^{-1} x=tan ^{-1} frac{2 x}{1-x^{2}`

Now,

`LHS=tan ^{-1} frac{2 tan frac{alpha}{2} tan (frac{pi}{4}-frac{beta}{2})}{1-tan ^{2} frac{alpha}{2} tan ^{2}(frac{pi}{4}-frac{beta}{2})}`

`=tan ^{-1} frac{tan frac{a}{2} frac{1-tan frac{beta}{2}}{1+tan frac{beta}{2}}}{1-tan ^{2} frac{a}{2}(frac{1-tan frac{beta}{2}}{1+tan frac{beta}{2}})^{1}}`

`=tan ^{-1} frac{2 tan frac{alpha}{2} cdot(1-tan ^{2} frac{beta}{2})}{(1+tan frac{beta}{2})^{2}-tan ^{2} frac{alpha}{2}(1-tan frac{beta}{2})^{2}}`

`=tan ^{-1} frac{2 tan frac{alpha}{2}(1-tan ^{2} frac{beta}{2})}{(1+tan ^{2} frac{beta}{2})(1-tan ^{2} frac{alpha}{2})+2 tan ^{frac{beta}{2}}(1+tan ^{2} frac{alpha}{2})}`

`=tan ^{-13}(frac{sin alpha cos beta}{cos alpha+sin beta})`

...