How do you prove $\left( {\tan x} \right)\left( {\cos x} \right) = \sin x$?

Hint:In the above question, the concept is based on the concept of trigonometric function. The main approach towards solving the above function is by knowing the tangent, sine and cosine functions can be solved by using the trigonometric identities on these functions and reducing it so that we can get a sine function on the right hand side.

Complete step by step solution:
Trigonometric function is also called circular functions and it defines as the function of the angle between the both sides. It tells us the relation between the angles and sides of the right-angle triangle.
There are three primary classification of functions i.e., sine, cosine, tangent. The other functions such as cotangent, secant and secant are derived from these primary functions.
So now we need to first look at the equation given,
$\left( {\tan x} \right)\left( {\cos x} \right) = \sin x$
We need to prove that the left hand side is equal to the right hand side.
So, on the left hand side we need to first simplify the equation. This can be done by converting tangent function into sine and cosine function. So below is the identity of tangent function.
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
No applying this in the equation we get,
$
\Rightarrow LHS = \left( {\tan x} \right)\left( {\cos x} \right) \\
\Rightarrow LHS = \left( {\dfrac{{\sin x}}{{\cos x}}} \right) \times \cos x \\
$
Since cosine function is common, we can cancel it and we get sine function.
\[LHS = \sin x\]
Hence the above equation is proved.

Note: An important thing to note is that sine function and cosine function always has a relation with tangent function because the formula of tangent function is \[\tan = \dfrac{{opposite}}{{adjacent}} = \dfrac{{\dfrac{{opposite}}{{hypotenuse}}}}{{\dfrac{{adjacent}}{{hypotenuse}}}} = \dfrac{{\sin
x}}{{\cos x}}\]
Since hypotenuse gets cancelled we get sine and cosine according to the formula.