Question

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

Answer 1

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.