Prove \[2{\tan ^{ - 1}}\left[ {\left( {\tan 45^\circ - \alpha } \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right] = {\cos ^{ - 1}}\left( {\dfrac{{\sin 2\alpha + \cos \beta }}{{1 + \sin 2\alpha \cos \beta }}} \right)\] .

Hint:To solve the question given above, we will use the properties of trigonometry. Different formulas of trigonometry will be used such as the formula for \[\tan \left( {a - b} \right)\] , and the formula for \[2{\tan ^{ - 1}}x\] . You also need to remember the value of \[\tan 45^\circ \] . The question needs to be done step wise in a detailed manner.

Formula used: The trigonometric formulas that will be used while solving the above question are:
The first formula is: \[\tan \left( {a - b} \right) = \left( {\dfrac{{\tan a - \tan b}}{{\tan a + \tan b}}} \right)\] .
The second formula is: \[2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\] .
The third formula is: \[\cos 2x = 1 - 2{\cos ^2}x\] .
The fourth formula is: \[\cos 2x = 2{\sin ^2}x - 1\] .

Complete step by step solution:
We have to prove: \[2{\tan ^{ - 1}}\left[ {\left( {\tan 45^\circ - \alpha } \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right] = {\cos ^{ - 1}}\left( {\dfrac{{\sin 2\alpha + \cos \beta }}{{1 + \sin 2\alpha \cos \beta }}} \right)\] .
We will solve the left-hand side first,
We have \[2{\tan ^{ - 1}}\left[ {\left( {\tan 45^\circ - \alpha } \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right]\] ,
By using the first formula, \[\tan \left( {a - b} \right) = \left( {\dfrac{{\tan a - \tan b}}{{\tan a + \tan b}}} \right)\] , we can write this as:
\[
  2{\tan ^{ - 1}}\left[ {\left( {\dfrac{{\tan 45^\circ - \tan \alpha }}{{\tan 45^\circ + \tan \alpha }}} \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right] \\
   \Rightarrow 2{\tan ^{ - 1}}\left[ {\left( {\dfrac{{1 - \tan \alpha }}{{1 + \tan \alpha }}} \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right] \\
 \]
Now we know that, \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] , putting this value of \[\tan x\] in the above equation, we get,
\[ \Rightarrow 2{\tan ^{ - 1}}\left[ {\left( {\dfrac{{\cos \alpha - \sin \alpha }}{{\cos \alpha + \sin \alpha }}} \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right]\] .
Now, to further evaluate this equation we will use the second formula \[2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\] .
Using this, we get,

\[
  {\cos ^{ - 1}}\left[ {\dfrac{{1 - {{\left\{ {\left( {\dfrac{{\cos \alpha - \sin \alpha }}{{\cos \alpha + \sin \alpha }}} \right)\tan \dfrac{\beta }{2}} \right\}}^2}}}{{1 + {{\left\{ {\left( {\dfrac{{\cos \alpha - \sin \alpha }}{{\cos \alpha + \sin \alpha }}} \right)\tan \dfrac{\beta }{2}} \right\}}^2}}}} \right] \\
   \Rightarrow {\cos ^{ - 1}}\left[ {\dfrac{{{{\left( {\cos \alpha + \sin \alpha } \right)}^2}{{\cos }^2}\dfrac{\beta }{2} - {{\left( {\cos \alpha - \sin \alpha } \right)}^2}{{\sin }^2}\dfrac{\beta }{2}}}{{{{\left( {\cos \alpha + \sin \alpha } \right)}^2}{{\cos }^2}\dfrac{\beta }{2} + {{\left( {\cos \alpha - \sin \alpha } \right)}^2}{{\sin }^2}\dfrac{\beta }{2}}}} \right] \\
 \]
On evaluating the above equation and using the formulas 3) \[\cos 2x = 1 - 2{\cos ^2}x\] and 4)\[\cos 2x = 2{\sin ^2}x - 1\] , it can be written as:
\[{\cos ^{ - 1}}\left[ {\dfrac{{\left( {1 + \sin 2\alpha } \right)\left( {\dfrac{{1 + \cos \beta }}{2}} \right) - \left( {1 - \sin 2\alpha } \right)\left( {\dfrac{{1 - \cos \beta }}{2}} \right)}}{{\left( {1 + \sin \alpha } \right)\left( {\dfrac{{1 + \cos \beta }}{2}} \right) + \left( {1 - \sin \alpha } \right)\left( {\dfrac{{1 - \cos \beta }}{2}} \right)}}} \right]\]
Now multiply the brackets.
\[{\cos ^{ - 1}}\left[ {\dfrac{{\dfrac{1}{2}\left\{ {\left( {1 + \sin 2\alpha + \cos \beta + \sin 2\alpha \cos \beta } \right) - \left( {1 - \sin 2\alpha - \cos \beta + \sin 2\alpha \cos \beta } \right)} \right\}}}{{\dfrac{1}{2}\left\{ {\left( {1 + \sin 2\alpha + \cos \beta + \sin 2\alpha \sin \beta } \right) + \left( {1 - \sin 2\alpha - \cos \beta + \sin 2\alpha \cos \beta } \right)} \right\}}}} \right]\]
On opening all the brackets, we get,
\[{\cos ^{ - 1}}\left[ {\dfrac{{1 - \sin 2\alpha + \cos \beta + \sin 2\alpha \cos \beta - 1 + \sin 2\alpha + \cos \beta - \sin 2\alpha \cos \beta }}{{1 + \sin 2\alpha + \cos \beta + \sin 2\alpha \cos \beta + 1 - \sin 2\alpha - \cos \beta + \sin 2\alpha \cos \beta }}} \right]\]
\[ \Rightarrow {\cos ^{ - 1}}\left[ {\dfrac{{2\left( {\sin 2\alpha + \cos \beta } \right)}}{{2 + 2\sin 2\alpha \cos \beta }}} \right]\]
We can divide 2 from both numerator and denominator, we get,
\[{\cos ^{ - 1}}\left[ {\dfrac{{\sin 2\alpha + \cos \beta }}{{1 + \sin 2\alpha \cos \beta }}} \right]\] which is equal to the right-hand side.
Hence, \[L.H.S = R.H.S\] .
So, we have proved that \[2{\tan ^{ - 1}}\left[ {\left( {\tan 45^\circ - \alpha } \right)\left( {\tan \dfrac{\beta }{2}} \right)} \right] = {\cos ^{ - 1}}\left( {\dfrac{{\sin 2\alpha + \cos \beta }}{{1 + \sin 2\alpha \cos \beta }}} \right)\] .

Note: Always remember the trigonometric formulas while solving questions similar to the above mentioned. These formulas will help you in solving the questions easily in very less amount of time. In the above question we have used four such trigonometric formulas. To prevent any errors, solve the question step by step.