
Find the value of $\tan \left( {\dfrac{{5\pi }}{3}} \right)$.
Answer
557.1k+ views
Hint:The Cartesian system divides the plane into 4 different quadrants.
Quadrant I: $0\; - \;\dfrac{\pi }{2}$
Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $
Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$
Quadrant IV: $\dfrac{{3\pi }}{2}\; - \;2\pi $
So in order to find the value of $\tan \left( {\dfrac{{5\pi }}{3}} \right)$ we first must convert $\left( {\dfrac{{5\pi }}{3}} \right)$as the sum of two numbers such that one of them forms the boundary of any quadrant and the other value acts as a reference value whose tan value is known to us.
Complete step by step solution:
Given
$\tan \left( {\dfrac{{5\pi }}{3}} \right)\;....................\left( i \right)$
We know that we have to convert $\left( {\dfrac{{5\pi }}{3}} \right)$as the sum of two numbers as mentioned above. So converting$\left( {\dfrac{{5\pi }}{3}} \right)$:
$\left( {\dfrac{{5\pi }}{3}} \right) = \left( {\pi \; + \;\dfrac{{2\pi }}{3}} \right)\;...............\left( {ii}
\right)$
Here $\pi $is a boundary of the quadrant also we can find the value of $\tan \left( {\dfrac{{2\pi }}{3}} \right)$$ \Rightarrow \tan \left( {\dfrac{{5\pi }}{3}} \right)\; = \;\tan \left( {\pi + \dfrac{{2\pi }}{3}} \right) = \tan \left( {\dfrac{{2\pi }}{3}} \right)$
Now converting $\left( {\dfrac{{2\pi }}{3}} \right)$ in the same manner as that of $\left( {\dfrac{{5\pi }}{3}} \right)$.
$ \Rightarrow \left( {\dfrac{{2\pi }}{3}} \right)\; = \;\left( {\pi - \dfrac{\pi }{3}}
\right).....................\left( {iii} \right)$
So here $\pi $is a boundary of the quadrant, also know the value of $\tan \left( {\dfrac{\pi }{3}} \right)$
$ \Rightarrow \tan \left( {\dfrac{{2\pi }}{3}} \right)\; = \tan \left( {\pi - \dfrac{\pi }{3}} \right)\; = \, -
\tan \left( {\dfrac{\pi }{3}} \right).............\left( {iv} \right)$
Here the negative sign occurs since $\left( {\pi - \dfrac{\pi }{3}} \right)$ is the Quadrant II and tan is negative in the Quadrant II.
So:
\[\tan \left( {\dfrac{{5\pi }}{3}} \right)\; = \; - \tan \left( {\dfrac{\pi }{3}} \right)\; = \; - \sqrt 3 = \; -
1.732050808...........\left( v \right)\]
Hence our final answer is given in $\left( v \right)$which is $ - \sqrt 3 = \; - 1.732050808$
Therefore: $\tan \left( {\dfrac{{5\pi }}{3}} \right)\; = - \sqrt 3 = \; - 1.732050808$
Note: tan is the abbreviation of tangent.
Also:
Quadrant I: $0\; - \;\dfrac{\pi }{2}$ All values are positive.
Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive.
Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.
Quadrant IV :$\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive.
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.
Quadrant I: $0\; - \;\dfrac{\pi }{2}$
Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $
Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$
Quadrant IV: $\dfrac{{3\pi }}{2}\; - \;2\pi $
So in order to find the value of $\tan \left( {\dfrac{{5\pi }}{3}} \right)$ we first must convert $\left( {\dfrac{{5\pi }}{3}} \right)$as the sum of two numbers such that one of them forms the boundary of any quadrant and the other value acts as a reference value whose tan value is known to us.
Complete step by step solution:
Given
$\tan \left( {\dfrac{{5\pi }}{3}} \right)\;....................\left( i \right)$
We know that we have to convert $\left( {\dfrac{{5\pi }}{3}} \right)$as the sum of two numbers as mentioned above. So converting$\left( {\dfrac{{5\pi }}{3}} \right)$:
$\left( {\dfrac{{5\pi }}{3}} \right) = \left( {\pi \; + \;\dfrac{{2\pi }}{3}} \right)\;...............\left( {ii}
\right)$
Here $\pi $is a boundary of the quadrant also we can find the value of $\tan \left( {\dfrac{{2\pi }}{3}} \right)$$ \Rightarrow \tan \left( {\dfrac{{5\pi }}{3}} \right)\; = \;\tan \left( {\pi + \dfrac{{2\pi }}{3}} \right) = \tan \left( {\dfrac{{2\pi }}{3}} \right)$
Now converting $\left( {\dfrac{{2\pi }}{3}} \right)$ in the same manner as that of $\left( {\dfrac{{5\pi }}{3}} \right)$.
$ \Rightarrow \left( {\dfrac{{2\pi }}{3}} \right)\; = \;\left( {\pi - \dfrac{\pi }{3}}
\right).....................\left( {iii} \right)$
So here $\pi $is a boundary of the quadrant, also know the value of $\tan \left( {\dfrac{\pi }{3}} \right)$
$ \Rightarrow \tan \left( {\dfrac{{2\pi }}{3}} \right)\; = \tan \left( {\pi - \dfrac{\pi }{3}} \right)\; = \, -
\tan \left( {\dfrac{\pi }{3}} \right).............\left( {iv} \right)$
Here the negative sign occurs since $\left( {\pi - \dfrac{\pi }{3}} \right)$ is the Quadrant II and tan is negative in the Quadrant II.
So:
\[\tan \left( {\dfrac{{5\pi }}{3}} \right)\; = \; - \tan \left( {\dfrac{\pi }{3}} \right)\; = \; - \sqrt 3 = \; -
1.732050808...........\left( v \right)\]
Hence our final answer is given in $\left( v \right)$which is $ - \sqrt 3 = \; - 1.732050808$
Therefore: $\tan \left( {\dfrac{{5\pi }}{3}} \right)\; = - \sqrt 3 = \; - 1.732050808$
Note: tan is the abbreviation of tangent.
Also:
Quadrant I: $0\; - \;\dfrac{\pi }{2}$ All values are positive.
Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive.
Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.
Quadrant IV :$\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive.
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

