Show that $\tan \left( 52.5{}^\circ \right)=\sqrt{6}-\sqrt{3}-\sqrt{2}+2$?

Hint: We will solve this question by using the trigonometric formulas of different functions. We will first consider the LHS of the given expression and then simplify it to prove it equal to RHS. We will first convert the tangent function in terms of sine and cosine function then by applying the trigonometric formulas we will get the desired answer. We will use following formulas of trigonometry to solve this question:
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
$\sin 2\theta =2\sin \theta \cos \theta $
$2{{\cos }^{2}}\theta =1+\cos 2\theta $

Complete step by step solution:
We have been given an expression $\tan \left( 52.5{}^\circ \right)=\sqrt{6}-\sqrt{3}-\sqrt{2}+2$.
We have to show that the given expression is true.
Let us consider the LHS of the given expression. Then we will get
$\Rightarrow \tan \left( 52.5{}^\circ \right)$
Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Now, applying the above formula to the given expression we will get
$\begin{align}
  & \Rightarrow \dfrac{\sin \left( 52.5{}^\circ \right)}{\cos \left( 52.5{}^\circ \right)} \\
 & \Rightarrow \dfrac{\sin \left( \dfrac{105{}^\circ }{2} \right)}{\cos \left( \dfrac{105{}^\circ }{2} \right)} \\
\end{align}$
Now, multiplying and dividing the above obtained equation by $\cos \left( \dfrac{105{}^\circ }{2} \right)$ and 2 we will get
$\Rightarrow \dfrac{2\sin \left( \dfrac{105{}^\circ }{2} \right)}{2\cos \left( \dfrac{105{}^\circ }{2} \right)}\times \dfrac{\cos \left( \dfrac{105{}^\circ }{2} \right)}{\cos \left( \dfrac{105{}^\circ }{2} \right)}$
Now, we now that $\sin 2\theta =2\sin \theta \cos \theta $ and $2{{\cos }^{2}}\theta =1+\cos 2\theta $
So by applying the above identities to the obtained equation we will get
$\Rightarrow \dfrac{\sin \left( \dfrac{105{}^\circ }{2}\times 2 \right)}{1+\cos \left( \dfrac{105{}^\circ }{2}\times 2 \right)}$
Now, simplifying the above obtained equation we will get
$\Rightarrow \dfrac{\sin \left( 105{}^\circ \right)}{1+\cos \left( 105{}^\circ \right)}$
Now, we can further simplifying the above obtained equation as
$\Rightarrow \dfrac{\sin \left( 60{}^\circ +45{}^\circ \right)}{1+\cos \left( 60{}^\circ +45{}^\circ \right)}$
Now, we know that $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$ and $\cos (A+B)=\cos A\cos B-\sin A\sin B$
Now, applying the formulas to the above obtained equation we will get
$\Rightarrow \dfrac{\sin 60{}^\circ \cos 45{}^\circ +\sin 45{}^\circ \cos 60{}^\circ }{1+\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ }$
Now, by trigonometric ratio table we get the values of all trigonometric functions. Then by substituting the values we will get
\[\Rightarrow \dfrac{\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}}{1+\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}}{1+\dfrac{1}{2\sqrt{2}}-\dfrac{\sqrt{3}}{2\sqrt{2}}} \\
 & \Rightarrow \dfrac{\dfrac{\sqrt{3}+1}{2\sqrt{2}}}{\dfrac{2\sqrt{2}+1-\sqrt{3}}{2\sqrt{2}}} \\
 & \Rightarrow \dfrac{\sqrt{3}+1}{2\sqrt{2}+1-\sqrt{3}} \\
\end{align}\]
Rationalising the denominator, we get
\[\begin{align}
 & \Rightarrow \dfrac{\left( \sqrt{3}+1 \right)\left( 2\sqrt{2}+1+\sqrt{3} \right)}{\left( 2\sqrt{2}+1-\sqrt{3} \right)\left( 2\sqrt{2}+1+\sqrt{3} \right)} \\
 & \Rightarrow \dfrac{\left( \sqrt{3}+1 \right)\left( 2\sqrt{2}+1+\sqrt{3} \right)}{{{\left( 2\sqrt{2}+1 \right)}^{2}}{{\left( \sqrt{3} \right)}^{2}}} \\
 & \Rightarrow \dfrac{\sqrt{3}+2\sqrt{6}+3+1+2\sqrt{2}+\sqrt{3}}{1+4\sqrt{2}+8-3} \\
 & \Rightarrow \dfrac{2\sqrt{3}+2\sqrt{6}+4+2\sqrt{2}}{6+4\sqrt{2}} \\
 & \Rightarrow \dfrac{2\left( \sqrt{3}+\sqrt{6}+2+\sqrt{2} \right)}{2\left( 3+2\sqrt{2} \right)} \\
\end{align}\]
Again rationalising the denominator, we get
\[\begin{align}
 & \Rightarrow \dfrac{\left( \sqrt{3}+\sqrt{6}+2+\sqrt{2} \right)\left( 3-2\sqrt{2} \right)}{\left( 3+2\sqrt{2} \right)\left( 3-2\sqrt{2} \right)} \\
 & \Rightarrow \dfrac{\sqrt{6}-\sqrt{3}+2-\sqrt{2}}{9-8} \\
 & \Rightarrow \sqrt{6}-\sqrt{3}-\sqrt{2}+2 \\
\end{align}\]
=RHS
Hence proved

Note: Here in this question we need to substitute the trigonometric identities and formulas. The solution is lengthy so be careful while substituting the values and solving the equations. Alternatively we can use the tangent formula given by $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ to simplify the LHS and get it equal to RHS.