Question

How do you use the angle sum identities to find the exact values of $\tan \left( {15} \right)$ and $\tan \left( {45} \right)$?

Answer 1

Hint: To solve this problem, you should know the formula for the addition and difference identity and also you must know how we can divide the $\theta $ into the addition form or in the difference form so that we can apply the value in either one of these identities and solve this problem.
Formula used: $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$

Complete answer:
To solve this question, we need to divide $\theta = 15$ into known $\theta $ values, so we have to divide $\theta = 15$ into \[\left( {45 - 30} \right)\] we get,
\[\tan {15^ \circ } = \tan ({45^ \circ } - {30^ \circ })\]
The formula for the difference identity is,
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
\[tan(45 - 30) = \dfrac{{tan45 - tan30}}{{1 + tan45\,tan30}}\]
It is mandatory to divide the value of $\theta $ into known values and now we can able to solve this problem as we know the value of\[\tan {45^ \circ } = 1\] , \[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\]. Applying the values in the formula we get,
\[tan(45 - 30) = \dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + 1\left( {\dfrac{1}{{\sqrt 3 }}} \right)}}\]
Now taking LCM in numerator as well as denominator, we get
$ \Rightarrow \tan (15) = \dfrac{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}$
On further simplification, we get
$ \Rightarrow \tan \left( {15} \right) = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
Now, we will calculate the value of $\tan \left( {{{30}^ \circ }} \right)$ using the angle difference identity for tangent as \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\].
So, we take angle $A = {45^ \circ }$ and $B = {15^ \circ }$.
Now, substituting the angles in the formulae, we get,
\[ \Rightarrow \tan \left( {{{45}^ \circ } - {{15}^ \circ }} \right) = \dfrac{{\tan {{45}^ \circ } - \tan {{15}^ \circ }}}{{1 + \tan {{45}^ \circ }\tan {{15}^ \circ }}}\]
Substituting the value of tangent of both the angles as $\tan {45^ \circ } = 1$ and \[\tan {15^ \circ } = \left( {\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)\], we get,
\[ \Rightarrow \tan \left( {{{30}^ \circ }} \right) = \dfrac{{1 - \left( {\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)}}{{1 + \left( 1 \right)\left( {\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)}}\]
Simplifying the calculations, we get,
\[ \Rightarrow \tan \left( {{{30}^ \circ }} \right) = \dfrac{{\left( {\dfrac{{\sqrt 3 + 1 - \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 + 1}}} \right)}}{{\left( {\dfrac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)}}\]
Cancelling the like terms with opposite signs, we get,
\[ \Rightarrow \tan \left( {{{30}^ \circ }} \right) = \dfrac{{\left( {\dfrac{2}{{\sqrt 3 + 1}}} \right)}}{{\left( {\dfrac{{2\sqrt 3 }}{{\sqrt 3 + 1}}} \right)}}\]
Cancelling the common factors, we get,
\[ \Rightarrow \tan \left( {{{30}^ \circ }} \right) = \dfrac{1}{{\sqrt 3 }}\]

Note:
We can also use the addition identity to solve this problem and the formula for addition identity is, $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ where we will add forty- five degree with fifteen degrees to get the sum of sixty degrees. In this identity we know the value of $\tan 45\,\,and\,\,\tan 60$ and therefore we will get the value of $\tan 15$.