How do you use the half angle formulas to determine the exact values of sine, cosine and tangent of the angle $\dfrac{7\pi }{12}$.

Hint: In this problem we need to calculate the sine, cosine and tangent values of the given angle by using the half angle formulas. To use the half angle formulas, we need to identify the quadrant in which the given angle lies. After that need to find the value of $\theta $ such that the cosine of the $\theta $ is known and the relation between the $\theta $ and given angle is a multiple of $2$.Now we will use the half angle formulas $\sin \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1-\cos \theta }{2}}$, $\cos \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1+\cos \theta }{2}}$ and $\tan \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}$.

Complete step by step answer:
Given angle $\dfrac{7\pi }{12}$.
We know that the angle $\dfrac{7\pi }{12}=105{}^\circ $ lies in the second quadrant. So, the sine value of the given angle is positive and the remaining values that means cosine and tangent values are negative.
Let us consider the angle $\dfrac{7\pi }{6}=210{}^\circ $.
We know the sine, cosine and tangent values for $\dfrac{7\pi }{6}$.
Now we have the relation between the angles $\dfrac{7\pi }{12}$, $\dfrac{7\pi }{6}$ as
$\begin{align}
  & \Rightarrow \dfrac{7\pi }{6}=2\times \dfrac{7\pi }{12} \\
 & \Rightarrow \dfrac{7\pi }{12}=\dfrac{\dfrac{7\pi }{6}}{2} \\
\end{align}$
From the above equation we can say that the considered angle and given angle has a relation in terms of $2$. So, we can use the angle $\dfrac{7\pi }{6}$ to calculate the value of $\dfrac{7\pi }{12}$.
Now the sine value of the given angle will be given by
$\Rightarrow \sin \left( \dfrac{7\pi }{12} \right)=\sqrt{\dfrac{1-\cos \dfrac{7\pi }{6}}{2}}$
We know that the value of $\cos \dfrac{7\pi }{6}=-\dfrac{\sqrt{3}}{2}$. Substituting this value in the above equation, then we will get
$\begin{align}
  & \Rightarrow \sin \left( \dfrac{7\pi }{12} \right)=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{2}} \\
 & \Rightarrow \sin \left( \dfrac{7\pi }{12} \right)=\sqrt{\dfrac{\dfrac{2+\sqrt{3}}{2}}{2}} \\
 & \Rightarrow \sin \left( \dfrac{7\pi }{12} \right)=\dfrac{\sqrt{2+3}}{2} \\
\end{align}$
Now the cosine value of the angle $\dfrac{7\pi }{12}$ is given by
$\begin{align}
  & \Rightarrow \cos \left( \dfrac{7\pi }{12} \right)=-\sqrt{\dfrac{1+\left( -\dfrac{\sqrt{3}}{2} \right)}{2}} \\
 & \Rightarrow \cos \left( \dfrac{7\pi }{12} \right)=-\sqrt{\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}} \\
 & \Rightarrow \cos \left( \dfrac{7\pi }{12} \right)=-\dfrac{\sqrt{2-3}}{2} \\
\end{align}$
Now the tangent value of the angle $\dfrac{7\pi }{12}$ is given by
$\begin{align}
  & \Rightarrow \tan \left( \dfrac{7\pi }{12} \right)=-\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{1+\left( -\dfrac{\sqrt{3}}{2} \right)}} \\
 & \Rightarrow \tan \left( \dfrac{7\pi }{12} \right)=-\sqrt{\dfrac{\dfrac{2+\sqrt{3}}{2}}{\dfrac{2-\sqrt{3}}{2}}} \\
 & \Rightarrow \tan \left( \dfrac{7\pi }{12} \right)=-\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}} \\
\end{align}$

Note:
For this type of problem students directly use the formulas and then write the solutions without considering the signs. But one should take care of the signs of the trigonometric ratios in their solutions. So, we need to consider the quadrant of the angle in which it lies.