Question

How do you find the exact value of \[\tan \dfrac{{5\pi }}{{12}}\]?

Answer 1

Hint: Here, we have to the exact value of the given trigonometric ratio. We will use the trigonometric complementary angles formula to find the suitable trigonometric ratio which is in relation to the given trigonometric ratio. Again by using a trigonometric identity, we will find a quadratic equation and by solving the quadratic equation, we will find the value of the given trigonometric ratio.

Formula Used:
We will use the following formulas:
Trigonometric complementary angles \[\tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \left( \theta \right)\]
Trigonometric Co-ratio \[\cot \theta = \dfrac{1}{{\tan \theta }}\]
Trigonometric Ratio \[\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}\]
Quadratic roots is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
The difference between the square numbers is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].

Complete Step by Step Solution:
We are given with a trigonometric ratio \[\tan \dfrac{{5\pi }}{{12}}\].
Now, we will rewrite the trigonometric ratio, so we get
\[\tan \dfrac{{5\pi }}{{12}} = \tan \left( {\dfrac{{6\pi }}{{12}} - \dfrac{\pi }{{12}}} \right)\]
Simplifying the above equation, we get
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \tan \left( {\dfrac{\pi }{2} - \dfrac{\pi }{{12}}} \right)\]
We know that Trigonometric complementary angles \[\tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \left( \theta \right)\]
By using the Trigonometric complementary angles, we get
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \cot \left( {\dfrac{\pi }{{12}}} \right)\]
We know that Trigonometric Co-ratio, \[\cot \theta = \dfrac{1}{{\tan \theta }}\]
By using the Trigonometric Co-ratio, we get
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \dfrac{1}{{\tan \left( {\dfrac{\pi }{{12}}} \right)}}\]………………………………………………………….\[\left( 1 \right)\]
Let us consider \[\tan \dfrac{\pi }{{12}} = \tan t\]
Now, we will find \[\tan 2t\] from\[\tan t\] .
\[\tan 2t = \tan \left( {2 \times \dfrac{\pi }{{12}}} \right)\]
By cancelling the terms, we get
\[ \Rightarrow \tan 2t = \tan \left( {\dfrac{\pi }{6}} \right)\]
Trigonometric Identity: \[\tan 2t = \dfrac{{2\tan t}}{{1 - {{\tan }^2}t}}\]
Now, by using the trigonometric identity, we get
\[ \Rightarrow \tan \dfrac{\pi }{6} = \dfrac{{2\tan t}}{{1 - {{\tan }^2}t}}\]
Now using the value \[\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}\] in the above equation, we get
\[ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{2\tan t}}{{1 - {{\tan }^2}t}}\]
On cross multiplication, we get
\[ \Rightarrow 1 - {\tan ^2}t = 2\sqrt 3 \tan t\]
\[ \Rightarrow {\tan ^2}t + 2\sqrt 3 \tan t - 1 = 0\]
Quadratic roots is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Comparing with the quadratic equation \[a{x^2} + bx + c = 0\] , we get \[a = 1,b = 2\sqrt 3\] and \[c = - 1\]
Now, we will solve the quadratic equation by using the roots, we get
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
By substituting the coefficient of \[{x^2}\], coefficient of \[x\] and the constant term, we get
\[ \Rightarrow \tan t = \dfrac{{ - 2\sqrt 3 \pm \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}\]
By simplifying the equation, we get
\[ \Rightarrow \tan t = \dfrac{{ - 2\sqrt 3 \pm \sqrt {12 + 4} }}{2}\]
Adding the terms, we get
\[ \Rightarrow \tan t = \dfrac{{ - 2\sqrt 3 \pm \sqrt {16} }}{2}\]
\[ \Rightarrow \tan t = \dfrac{{ - 2\sqrt 3 \pm 4}}{2}\]
By taking out the common factors, we get
\[ \Rightarrow \tan t = \dfrac{{2\left( { - \sqrt 3 \pm 2} \right)}}{2}\]
Cancelling the like terms, we get
\[ \Rightarrow \tan t = \left( { - \sqrt 3 \pm 2} \right)\]
\[ \Rightarrow \tan t = - \sqrt 3 \pm 2\]
We know that \[\tan \dfrac{\pi }{{12}} = \tan t\]
\[ \Rightarrow \tan \dfrac{\pi }{{12}} = - \sqrt 3 \pm 2\]
We know that \[\tan \dfrac{\pi }{{12}}\] is positive. So, we get
\[ \Rightarrow \tan \dfrac{\pi }{{12}} = 2 - \sqrt 3 \]
By substituting \[\tan \dfrac{\pi }{{12}} = 2 - \sqrt 3 \]in equation\[\left( 1 \right)\], we get
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \dfrac{1}{{2 - \sqrt 3 }}\]
Now, by multiplying the numerator and denominator with \[2 + \sqrt 3 \] which is the conjugate of \[2 - \sqrt 3 \].
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \dfrac{1}{{2 - \sqrt 3 }} \times \dfrac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}\]
The difference between the square numbers is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Now, by using the algebraic identity, we get
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \dfrac{{2 + \sqrt 3 }}{{{2^2} - {{\left( {\sqrt 3 } \right)}^2}}}\]
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = \dfrac{{2 + \sqrt 3 }}{{4 - 3}}\]
By simplifying the equation, we get
\[ \Rightarrow \tan \dfrac{{5\pi }}{{12}} = 2 + \sqrt 3 \]

Therefore, the value of \[\tan \dfrac{{5\pi }}{{12}}\] is \[2 + \sqrt 3 \].

Note:
We know that Trigonometric Equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation that is always true for all the variables. We should know that we have many trigonometric identities that are related to all the other trigonometric equations. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right-angled triangle with respect to any of its acute angle. Trigonometric Ratios are used to find the relationships between the sides of a right-angle triangle. Conjugate is a term where the sign is changed between two terms.