How do you verify the identity $ \tan \left( x+45 \right)=\dfrac{1+\tan x}{1-\tan x} $ ?

Hint: We have to first use the associative law of ratio tan. We use the formula of $ \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{\tan a-\tan b} $ to put the values of $ a=45,b=x $ . We can also verify it using arbitrary values as $ x=45 $ .

Complete step-by-step answer:
We have to verify the identity $ \tan \left( x+45 \right)=\dfrac{1+\tan x}{1-\tan x} $ by using the laws of associative angles.
We know that $ \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{\tan a-\tan b} $ .
We have to replace the values in the formula to verify the identity $ \tan \left( x+45 \right)=\dfrac{1+\tan x}{1-\tan x} $ .
We replace it with $ a=45,b=x $ in $ \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{\tan a-\tan b} $ .
Putting the values, we get $ \tan \left( 45+x \right)=\dfrac{\tan 45+\tan x}{\tan 45-\tan x} $ .
Now we know that the trigonometric ratio tan at the value of 45 gives $ \tan 45=1 $ .
We put the value and get
$ \tan \left( x+45 \right)=\dfrac{\tan 45+\tan x}{\tan 45-\tan x}=\dfrac{1+\tan x}{1-\tan x} $ .
Thus, proved that $ \tan \left( x+45 \right)=\dfrac{1+\tan x}{1-\tan x} $ .
We can also take an arbitrary value for $ x=45 $ .
We put the value in the expression of $ \tan \left( x+45 \right)=\dfrac{1+\tan x}{1-\tan x} $ .
The left-hand side of the expression becomes \[\tan \left( 45+45 \right)=\tan 90=\text{undefined}\].
The right-hand side of the expression becomes \[\dfrac{1+\tan x}{1-\tan x}=\dfrac{1+\tan 45}{1-\tan 45}=\dfrac{1+1}{1-1}=\text{undefined}\]
Thus, it is also verified.

Note: We need to remember that the additional value for the ratio tan comes from the associative rules of sin and cos. It is defined for any other values also.