What is the value of a trigonometric expression \[\dfrac{{\left( {\tan70^{\circ } - \tan20^{\circ }} \right)}}{{\tan50^{\circ }}}\]?
A. 2
B. 1
C. 0
D. 3

Hint: First we consider, \[\left( {70^{\circ } - 20^{\circ }} \right) = 50^{\circ }\]. Then take tan on both sides. On left-hand side apply the identity of \[\tan\left( {A - B} \right)\]. Further simplify the equation to reach the required answer.

Formula Used:
\[\tan\left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \tan B}}\]
\[\tan A = \dfrac{1}{{\cot A }}\]
\[\tan\left( {90^{\circ } - A} \right) = \cot A\]

Complete step by step solution:
The given trigonometric expression is \[\dfrac{{\left( {\tan70^{\circ } - \tan20^{\circ }} \right)}}{{\tan50^{\circ }}}\].
Let’s find out the value of the above expression.
Let consider,
\[\left( {70^{\circ } - 20^{\circ }} \right) = 50^{\circ }\]
Take \[\tan\] on both sides.
\[\tan\left( {70^{\circ } - 20^{\circ }} \right) = \tan50^{\circ }\]
Now apply the identity \[\tan\left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \tan B}}\] on the left-hand side of the equation.
\[\dfrac{{\tan70^{\circ } - \tan20^{\circ }}}{{1 + \tan70^{\circ }\tan20^{\circ }}} = \tan50^{\circ }\]
Simplify the above equation.
\[\dfrac{{\tan70^{\circ } - \tan20^{\circ }}}{{\tan50^{\circ }}} = 1 + \tan70^{\circ }\tan20^{\circ }\]
\[ \Rightarrow \]\[\dfrac{{\tan70^{\circ } - \tan20^{\circ }}}{{\tan50^{\circ }}} = 1 + \tan70^{\circ }\tan\left( {90^{\circ } - 70^{\circ }} \right)\]
Apply the identity \[\tan\left( {90^{\circ } - A} \right) = \cot A\] on the right-hand side of the equation.
\[\dfrac{{\tan70^{\circ } - \tan20^{\circ }}}{{\tan50^{\circ }}} = 1 + \tan70^{\circ }\cot70^{\circ }\]
\[ \Rightarrow \]\[\dfrac{{\tan70^{\circ } - \tan20^{\circ }}}{{\tan50^{\circ }}} = 1 + \tan70^{\circ }\left( {\dfrac{1}{{\tan70^{\circ }}}} \right)\] [ Since \[\tan A = \dfrac{1}{{\cot A }}\]]
\[ \Rightarrow \]\[\dfrac{{\tan70^{\circ} - \tan20^{\circ }}}{{\tan50^{\circ }}} = 1 + 1\]
\[ \Rightarrow \]\[\dfrac{{\tan70^{\circ } - \tan20^{\circ }}}{{\tan50^{\circ }}} = 2\]

Hence the correct option is A.

Note: Trigonometric expressions can be solved by converting the sum or differences of sine, cosine, or tangent functions to the product of trigonometric ratios.
Students often get confused with the formulas \[tan\left( {A - B} \right)\] and \[tan\left( {A + B} \right)\].
The correct formulas are:
\[tan\left( {A - B} \right) = \dfrac{{tanA - tanB}}{{1 + tanA tanB}}\]
\[tan\left( {A + B} \right) = \dfrac{{tanA + tanB}}{{1 - tanA tanB}}\]