How do you prove \[{\tan ^{ - 1}}\left( {\dfrac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]?

Hint: In this question, we have to prove the given trigonometric equations. We need to apply a trigonometric formula for inverse tangent functions. After that simplifying the value within the bracket we will get the value of the inverse tangent function as given in the question and we can prove the required result.
Trigonometric formula:
\[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\], where \[xy < 1\]
\[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\], where \[xy > 1\]

Complete step-by-step solution:
We need to prove, \[{\tan ^{ - 1}}\left( {\dfrac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\].
Left hand side =\[{\tan ^{ - 1}}\left( {\dfrac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right)\]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{4} + \dfrac{2}{9}}}{{1 - \dfrac{1}{4} \times \dfrac{2}{9}}}} \right)\][Taking \[x = \dfrac{1}{4} \& y = \dfrac{2}{9}\] and applying \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{9 + 8}}{{36}}}}{{1 - \dfrac{1}{{18}}}}} \right)\]
Simplifying we get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{17}}{{36}}}}{{\dfrac{{18 - 1}}{{18}}}}} \right)\]
Let us subtract the numerator term and we get
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{17}}{{36}}}}{{\dfrac{{17}}{{18}}}}} \right)\]
Taking it as reciprocal and we can write it as,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{17}}{{36}} \times \dfrac{{18}}{{17}}} \right)\]
Solving we get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{18}}{{36}}} \right)\]
Let us divide the term and we get
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
=Right Hand side
Hence, Left hand side = Right hand side.

Therefore we get, \[{\tan ^{ - 1}}\left( {\dfrac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\dfrac{2}{9}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\](Proved).

Note: sin cos formulas are based on sides of the right-angled triangle. Sin and Cos are basic trigonometric functions along with tan function, in trigonometry. Sine of angle is equal to the ratio of opposite side and hypotenuse whereas cosine of an angle is equal to ratio of adjacent side and hypotenuse.
\[\sin \theta = \dfrac{{Opposite\, side}}{{Hypotenuse}}\]
\[\cos \theta = \dfrac{{Adjacent\, side}}{{Hypotenuse}}\]

In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. The most widely used trigonometric functions are the sine, the cosine, and the tangent.
The inverse tangent functions \[{\sin ^{ - 1}}x,{\cos ^{ - 1}}x,{\tan ^{ - 1}}x\] are used to find the unknown measure of an angle of a right angled triangle when two side lengths are known.