Question

How do you use the angle difference identity to find the exact value of $\tan 75^\circ $?

Answer 1

Hint:
Start with writing down the formula for angle difference in tangent, i.e. $\tan \left( {A \pm B} \right) = \dfrac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}$ .Now you can express the angle such that $75 = A + B$ and values of $\tan A$ and $\tan B$ are already known, i.e. $75 = 30 + 45$. Now substitute the values and solve the right-hand side of the equation to get the required answer.

Complete step by step solution:
Here in this problem, we are given an expression $\tan 75^\circ $ and we need to find the value of this tangent function using the angle difference identity of trigonometric identities.
Before starting with the solution we must understand a few concepts of the tangent function and angle difference identity. Tangent is one of the trigonometric ratios which represent the ratio of the opposite side and the adjacent side to an angle in a right-angled triangle. The angle difference identity of the tangent function gives us the relation between two angles and can be written as:
$ \Rightarrow \tan \left( {A \pm B} \right) = \dfrac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}$ , where angle ‘A’ and ‘B’ are two the angles
In our case, we are given an angle of measure $75^\circ $ , which can be expressed as the difference of two angles as:
$ \Rightarrow 75^\circ = 30^\circ + 45^\circ $
Now using the angle difference formula by putting $A = 30^\circ {\text{ and }}B = 45^\circ $
$ \Rightarrow \tan \left( {30^\circ + 45^\circ } \right) = \dfrac{{\tan 30^\circ + \tan 45^\circ }}{{1 - \tan 30^\circ \tan 45^\circ }}$
As we know according to the definition of the tangent, the values of $\tan 30^\circ {\text{ and }}\tan 45^\circ $ can be written as:
$ \Rightarrow \tan 30^\circ = \dfrac{{\sin 30^\circ }}{{\cos 30^\circ }} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{1}{{\sqrt 3 }}$
And similarly;
$ \Rightarrow \tan 45^\circ = \dfrac{{\sin 45^\circ }}{{\cos 45^\circ }} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}} = 1$
Now we can substitute this value in the expression of angle difference identity.
$ \Rightarrow \tan \left( {75^\circ } \right) = \tan \left( {30^\circ + 45^\circ } \right) = \dfrac{{\tan 30^\circ + \tan 45^\circ }}{{1 - \tan 30^\circ \tan 45^\circ }} = \dfrac{{\dfrac{1}{{\sqrt 3 }} + 1}}{{1 - \dfrac{1}{{\sqrt 3 }} \times 1}}$
This above expression can be solved by simplifying numerator and denominator separately as:
$ \Rightarrow \tan \left( {75^\circ } \right) = \dfrac{{\dfrac{1}{{\sqrt 3 }} + 1}}{{1 - \dfrac{1}{{\sqrt 3 }} \times 1}} = \dfrac{{\dfrac{{1 + \sqrt 3 }}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$
The above-obtained solution for the value of $\tan 75^\circ $ can be rationalized into the simplest form
\[ \Rightarrow \tan \left( {75^\circ } \right) = \dfrac{{\left( {\sqrt 3 + 1} \right) \times \left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right) \times \left( {\sqrt 3 + 1} \right)}} = \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {\sqrt 3 - 1} \right) \times \left( {\sqrt 3 + 1} \right)}}\]
As we know the algebraic identity of the difference of square, i.e. ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ and the identity for the square of sum, i.e. ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ . This can be used in the denominator and numerator of the above fraction
\[ \Rightarrow \tan \left( {75^\circ } \right) = \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {\sqrt 3 - 1} \right) \times \left( {\sqrt 3 + 1} \right)}} = \dfrac{{\left( {{{\left( {\sqrt 3 } \right)}^2} + {1^2} + 2 \times \sqrt 3 \times 1} \right)}}{{\left( {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} \right)}} = \dfrac{{\left( {3 + 1 + 2\sqrt 3 } \right)}}{{\left( {3 - 1} \right)}} = \dfrac{{4 + 2\sqrt 3 }}{2}\]
After taking common the number $2$ from the numerator, we get:
\[ \Rightarrow \tan \left( {75^\circ } \right) = \dfrac{{2 \times \left( {2 + \sqrt 3 } \right)}}{2} = 2 + \sqrt 3 \]

Therefore, we get the required value of $\tan 75^\circ $ using the angle difference identity is $\tan 75 ^{\circ} = 2+ \sqrt{3}$

Note:
In this question, we expressed the given angle of $75^\circ $ into $30^\circ {\text{ and }}45^\circ $ because of the already known values for the tangent of these angles. Also notice that similar representation can be $75 = 90 - 15$ , but the value for the tangent of $90^\circ $ is not defined. An alternative approach to finding the value for the same angle can be the use of tan double angle identity and angle difference identity together. You can use $\tan \left( {75^\circ + 75^\circ } \right) = \tan 150^\circ = \tan \left( {180^\circ - 30^\circ } \right)$ and then solve it further to get the required value.