How do you prove $\dfrac{{1 + \tan x}}{{1 + \cot x}}$?

Hint:In order to simplify the above statement, rewrite the expression using the rule of trigonometry that $\cot x$is equal to the reciprocal of $\tan x$,and after taking the LCM in the denominator you’ll see that the numerator and denominator are both same ,so cancelling out both result in 1 and only $\tan x$is left with the expression which is the required answer.

Complete step by step solution:
We are given a trigonometric function$\dfrac{{1 + \tan x}}{{1 + \cot x}}$ which we have to simplify

Recall properties of trigonometry that $\cot x = \dfrac{1}{{\tan x}}$.So, now replacing $\cot x$with this in our function.
$ = \dfrac{{1 + \tan x}}{{1 + \dfrac{1}{{\tan x}}}}$

Taking LCM in the denominator
$ = \dfrac{{1 + \tan x}}{{\dfrac{{\tan x + 1}}{{\tan x}}}}$

Now rewriting the expression
$ = \dfrac{{1 + \tan x}}{{1 + \tan x}}.(\tan x)$

Cancelling out the numerator and denominator, it results into $1$
\[
= 1.(\tan x) \\
= \tan x = \dfrac{{\sin x}}{{\cos x}} \\
\]

Therefore, the implication of function$\dfrac{{1 + \tan x}}{{1 + \cot x}}$is equal to \[\tan
x\,or\,\dfrac{{\sin x}}{{\cos x}}\].

Formula:
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
$\cot x = \dfrac{1}{{\tan x}}$

Note:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.

2. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.

If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.

3. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.

Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.

We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta )
= - \tan \theta $