Question

The value of \[\tan 75^\circ \]is
(A) $1 + \dfrac{1}{{\sqrt 3 }}$
(B) $2 - \sqrt 3 $
(C) $2 + \sqrt 3 $
(D) $1 + \sqrt 3 $

Answer 1

Hint: The trigonometry is a concept in mathematics that deals with the sides, angles of a body. For suppose, if a body is tilted with some angle, then to find the angle in normal lengths or distance of that tilted body we use trigonometry and its formula. In this solution, split the angle \[75^\circ \] into two different angles like \[45^\circ \]and \[30^\circ \]. So we can substitute those values in the formula of trigonometry to find the solution.

Complete step-by-step answer:
Given:
The trigonometric function is \[\tan 75^\circ \].
Also, we can write the given value of \[\tan 75^\circ \]in the form of,
\[\tan 75^\circ = \tan \left( {45^\circ + 30^\circ } \right)\]
We know the formula to solve the value is given as,
\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
On substituting the values in the above equation, we get
\[\begin{array}{c}
\tan \left( {A + B} \right) = \dfrac{{\tan 45^\circ + \tan 30^\circ }}{{1 - \tan 45^\circ \tan 30^\circ }}\\
 = \dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 - 1.\dfrac{1}{{\sqrt 3 }}}}\\
 = \dfrac{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}\\
 = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}
\end{array}\]
Therefore, the value of \[\tan 75^\circ \]is \[\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\].

Note: Make sure about the trigonometry table, there will be different values for different angles in the trigonometry properties. While substituting such values, be careful and solve them, even a small value will result wrong. The relation between angles is addition so the formula consists of negative sign in the denominator and positive sign in the numerator, if the relation between the angle is subtraction then the formula consists of negative symbol in the numerator and positive symbol in the denominator.