Question

The value of $\tan 150^\circ = ?$
A.$\dfrac{1}{{\sqrt 3 }}$
B.$- \dfrac{1}{{\sqrt 3 }}$
C.$\sqrt 3$
D.$- \sqrt 3$

Answer 1

Hint : We cannot directly find the value of $\tan 150^\circ$. So, we can write 150 as 180 minus 30. That is $\tan \left( {\pi - 30^\circ } \right)$. Now, as 30 degree is being subtracted from $\pi$, $\tan \left( {\pi - 30^\circ } \right)$ will lie between $\dfrac{\pi }{2}$ and $\pi$, that is 2nd quadrant. We know that $\tan \theta$ is negative in the 2nd quadrant. So, the value of $\tan 150^\circ$ will be equal to the negative value of $\tan 30^\circ$.

Complete step-by-step answer :
In this question, we are asked to find the value of $\tan 150^\circ$. Now, we don’t have direct value for $\tan 150^\circ$. So, we need to use some trigonometric relations and formulas to find the value of $\tan 150^\circ$.
Now, we can write 150 as 180 minus 30. Therefore, we get
$\Rightarrow \tan 150^\circ = \tan \left( {180^\circ - 30^\circ } \right)$ - - - - - - - - - - - - - - - (1)
Now, we know that 180 degrees is equal to $\pi$. Therefore, equation (1) becomes
$\Rightarrow \tan 150^\circ = \tan \left( {\pi - 30^\circ } \right)$ - - - - - - - - - - - - - - - (2)
Now, as $30^\circ$ is being subtracted from $\pi$, $\tan \left( {\pi - 30^\circ } \right)$ will lie between $\dfrac{\pi }{2}$ and $\pi$, that is 2nd quadrant. We know that in the 2nd quadrant only sin and cosec are positive.
Hence, in the 2nd quadrant the value of $\tan \left( {\pi - 30^\circ } \right)$ will be negative.
$\Rightarrow \tan 150^\circ = \tan \left( {\pi - 30^\circ } \right)$
$= -\tan 30^\circ \\\ = -\dfrac{1}{{\sqrt 3 }} \\\ = - \dfrac{1}{{\sqrt 3 }} \;$
Hence, option B is the correct answer.
So, the correct answer is “Option B”.

Note : $\Rightarrow$1st quadrant = All positive
$\Rightarrow$2nd quadrant = $\sin \theta$ and $\cos ec\theta$ are positive.
$\Rightarrow$3rd quadrant = $\tan \theta$ and $\cot \theta$ are positive.
$\Rightarrow$4th quadrant = $\cos \theta$ and $\sec \theta$ are positive.