How do you evaluate $\tan (150)$.
Answer
575.7k+ views
Hint:
In order to solve such a type of sum with values not common for example $\tan (120)$ or above sum, the student has to make use of the formula of $\tan $ to evaluate. To evaluate the above sum the student can substitute the value directly or can use the formula $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \times \tan B}}$ as we know the value of $\tan (180)\& \tan (30)$. Another way to solve such a type of sum is by proving it.
Complete step by step solution:
In this Method the formula $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \times \tan B}}$ is to be used to find the answer for $\tan ({150^ \circ })$.
We can write $\tan ({150^ \circ })$ as $\tan ({180^ \circ } - {30^ \circ })$.
Using the formula :
$\tan (180 - 30) = \dfrac{{\tan {{180}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{180}^ \circ } \times \tan {{30}^ \circ }}}........(1)$
\[\tan (180 - 30) = \dfrac{{\dfrac{{\sin {{180}^ \circ }}}{{\cos {{180}^ \circ }}} - \dfrac{{\sin {{30}^ \circ }}}{{\cos {{30}^ \circ }}}}}{{1 + \dfrac{{\sin {{180}^ \circ }}}{{\cos {{180}^ \circ }}} \times \dfrac{{\sin {{30}^ \circ }}}{{\cos {{30}^ \circ }}}}}..........(2)\]
As we know that value of $\sin {180^ \circ }$ is $0$, We can say that the value of $\tan ({180^ \circ })$ is $0$
Substituting this value in equation $2$
\[\tan (180 - 30) = \dfrac{{0 - \dfrac{1}{{\sqrt 3 }}}}{1}..........(3)\]
Making use of the standard values, we get the value of $\tan ({150^ \circ })$
$\tan ({180^ \circ } - {30^ \circ }) = - \dfrac{1}{{\sqrt 3 }}.........(4)$
$\therefore \tan ({150^ \circ }) = - \dfrac{1}{{\sqrt 3 }}$
Note:
In order to solve trigonometric numerical the student should know the standard values of all the trigonometric functions which are ${0^ \circ }, {30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$. These values would not only be useful for trigonometric related problems but also help in calculating the values of sides in triangles, useful in the heights and distances chapter. Also they are useful while solving chapters like integration, derivatives. Thus students are advised to memorize the standard values of trigonometric functions.
In order to solve such a type of sum with values not common for example $\tan (120)$ or above sum, the student has to make use of the formula of $\tan $ to evaluate. To evaluate the above sum the student can substitute the value directly or can use the formula $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \times \tan B}}$ as we know the value of $\tan (180)\& \tan (30)$. Another way to solve such a type of sum is by proving it.
Complete step by step solution:
In this Method the formula $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \times \tan B}}$ is to be used to find the answer for $\tan ({150^ \circ })$.
We can write $\tan ({150^ \circ })$ as $\tan ({180^ \circ } - {30^ \circ })$.
Using the formula :
$\tan (180 - 30) = \dfrac{{\tan {{180}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{180}^ \circ } \times \tan {{30}^ \circ }}}........(1)$
\[\tan (180 - 30) = \dfrac{{\dfrac{{\sin {{180}^ \circ }}}{{\cos {{180}^ \circ }}} - \dfrac{{\sin {{30}^ \circ }}}{{\cos {{30}^ \circ }}}}}{{1 + \dfrac{{\sin {{180}^ \circ }}}{{\cos {{180}^ \circ }}} \times \dfrac{{\sin {{30}^ \circ }}}{{\cos {{30}^ \circ }}}}}..........(2)\]
As we know that value of $\sin {180^ \circ }$ is $0$, We can say that the value of $\tan ({180^ \circ })$ is $0$
Substituting this value in equation $2$
\[\tan (180 - 30) = \dfrac{{0 - \dfrac{1}{{\sqrt 3 }}}}{1}..........(3)\]
Making use of the standard values, we get the value of $\tan ({150^ \circ })$
$\tan ({180^ \circ } - {30^ \circ }) = - \dfrac{1}{{\sqrt 3 }}.........(4)$
$\therefore \tan ({150^ \circ }) = - \dfrac{1}{{\sqrt 3 }}$
Note:
In order to solve trigonometric numerical the student should know the standard values of all the trigonometric functions which are ${0^ \circ }, {30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$. These values would not only be useful for trigonometric related problems but also help in calculating the values of sides in triangles, useful in the heights and distances chapter. Also they are useful while solving chapters like integration, derivatives. Thus students are advised to memorize the standard values of trigonometric functions.
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