Show that $2{\tan ^{ - 1}}\left\{ {\tan \dfrac{\alpha }{2}\tan \left( {\dfrac{\pi }{4} - \dfrac{\beta }{2}} \right)} \right\} = {\tan ^{ - 1}}\left( {\dfrac{{\sin \alpha \cos \beta }}{{\cos \alpha + \sin \beta }}} \right)$

Hint: So to solve this question, two things are the most important first one is the use of proper formula and the second one is which side we take to proceed with the proof. Therefore, in this question, we will proceed with the proof by using LHS and one formula will be used there.

Formula used:
The trigonometric formula used in this question-
$2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$

Complete step-by-step answer:
So we have the trigonometric equation
$2{\tan ^{ - 1}}\left\{ {\tan \dfrac{\alpha }{2}\tan \left( {\dfrac{\pi }{4} - \dfrac{\beta }{2}} \right)} \right\} = {\tan ^{ - 1}}\left( {\dfrac{{\sin \alpha \cos \beta }}{{\cos \alpha + \sin \beta }}} \right)$
And we have to show that both of them should be equal.
So let’s take the LHS side and proceed with the equation.
By using the formula$2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$, we get
$ \Rightarrow {\tan ^{ - 1}}\dfrac{{2\tan \dfrac{\alpha }{2}.\tan \left( {\dfrac{\pi }{4} - \dfrac{\beta }{2}} \right)}}{{1 - {{\tan }^2}\dfrac{\alpha }{2}.{{\tan }^2}\left( {\dfrac{\pi }{4} - \dfrac{\beta }{2}} \right)}}$
And the degree of $\tan $ can also be written as
$ \Rightarrow {\tan ^{ - 1}}\dfrac{{2\tan \dfrac{\alpha }{2}\dfrac{{1 - \tan \dfrac{\beta }{2}}}{{1 + \tan \dfrac{\beta }{2}}}}}{{1 - {{\tan }^2}\dfrac{\alpha }{2}{{\left( {\dfrac{{1 - \tan \dfrac{\beta }{2}}}{{1 + \tan \dfrac{\beta }{2}}}} \right)}^2}}}$
Now on further solving more and expanding the equation, we get
\[ \Rightarrow {\tan ^{ - 1}}\dfrac{{2\tan \dfrac{\alpha }{2}\left( {1 - {{\tan }^2}\dfrac{\beta }{2}} \right)}}{{{{\left( {1 + \tan \dfrac{\beta }{2}} \right)}^2} - {{\tan }^2}\dfrac{\alpha }{2}{{\left( {1 - \tan \dfrac{\beta }{2}} \right)}^2}}}\]
Again solving, we get
\[ \Rightarrow {\tan ^{ - 1}}\dfrac{{2\tan \dfrac{\alpha }{2}\left( {1 - {{\tan }^2}\dfrac{\beta }{2}} \right)}}{{\left( {1 + {{\tan }^2}\dfrac{\beta }{2}} \right)\left( {1 - {{\tan }^2}\dfrac{\alpha }{2}} \right) + 2\tan \dfrac{\beta }{2}\left( {1 + {{\tan }^2}\dfrac{\alpha }{2}} \right)}}\]
Now by combining the term, and applying the formula we get
\[ \Rightarrow {\tan ^{ - 1}}\dfrac{{\dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}\dfrac{{1 - {{\tan }^2}\dfrac{\beta }{2}}}{{1 + {{\tan }^2}\dfrac{\beta }{2}}}}}{{\dfrac{{1 - {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}\dfrac{{2\tan \dfrac{\beta }{2}}}{{1 + {{\tan }^2}\dfrac{\beta }{2}}}}}\]
And it can also be written as,
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\sin \alpha \cos \beta }}{{\cos \alpha + \sin \beta }}} \right)$
And it will be equal to the right-hand side and this is proved that the RHS is equal to the LHS.
So we can say that $2{\tan ^{ - 1}}\left\{ {\tan \dfrac{\alpha }{2}\tan \left( {\dfrac{\pi }{4} - \dfrac{\beta }{2}} \right)} \right\} = {\tan ^{ - 1}}\left( {\dfrac{{\sin \alpha \cos \beta }}{{\cos \alpha + \sin \beta }}} \right)$.

Note: This question can also be proved by taking the right-hand side and by using the different formulas we can prove. But we should always try to use the most convenient and easiest method to prove. As we can see that we had easily shown how the left-hand side becomes equal to the right-hand side. And we know when to apply the formula and which formula. And we should always try to use only sine and cosine terms as it makes it easier to solve the functions. And the last one I would suggest is the practice. Only this can give accuracy.