Question

How do you prove $\tan \left[ {x + \left( {\dfrac{\pi }{4}} \right)} \right] = \dfrac{{1 + \tan x}}{{1 - \tan x}}?$

Answer 1

For solving this question, we just need knowledge of some basic trigonometric formulas and relations . We are going to use two simple formulas .
One is the property of tangent which is $\tan \left( {\dfrac{\pi }{4}} \right) = 1$
And the other one is $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ .
First of all, we will be substituting $1$ with $\tan \left( {\dfrac{\pi }{4}} \right)$ and then adjust the terms in such a way that can compare the given equation with the above formula

Complete step by step answer:
In this question, we are supposed to simplify $\tan \left[ {x + \left( {\dfrac{\pi }{4}} \right)} \right]$
To solve this question, we are going to make some modifications in the above equation.
Comparing the above formula we get $A = x$ and $B = \dfrac{\pi }{4}$ .
Use the formula in the above function and we get
L.H.S.,
$\tan \left[ {x + \left( {\dfrac{\pi }{4}} \right)} \right] = \dfrac{{\tan x + \tan \left( {\dfrac{\pi }{4}} \right)}}{{1 - \tan x\tan \left( {\dfrac{\pi }{4}} \right)}}$
First of all , we know that the value of $\tan \left( {\dfrac{\pi }{4}} \right) = 1$
So, substitute $\tan \left( {\dfrac{\pi }{4}} \right)$ in above equation with $1$ and we get
$= \dfrac{{\tan x + 1}}{{1 - \tan x \times 1}}$
$= \dfrac{{1 + \tan x}}{{1 - \tan x}}$
$= R.H.S.$
Therefore we get $\tan \left[ {x + \left( {\dfrac{\pi }{4}} \right)} \right] = \dfrac{{1 + \tan x}}{{1 - \tan x}}$ [proved]
Therefore the required condition is satisfied.

Note:
Trigonometric questions are always formula based. So, remember that you need to learn each and every trigonometric formula and always keep them in mind while solving trigonometric questions. Sometimes you have to use relations between them too to substitute the values in the question and get the solution. Here we substitute the value of $\tan \left( {\dfrac{\pi }{4}} \right)$ , which should be $1$ . Now we explain how to get it. We know that $\tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\sin \left( {\dfrac{\pi }{4}} \right)}}{{\cos \left( {\dfrac{\pi }{4}} \right)}}$ and we know the value of $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} = \cos \left( {\dfrac{\pi }{4}} \right)$ . We put this in the above equation and we get $\tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}} = 1$ .