How do you find the value of $\tan \dfrac{\theta }{2}$ given that $\cos \theta = - \dfrac{4}{5}$ and $\theta $ is in quadrant second?

Hint:
In the given question we have to find the value of half angle of tan function in the trigonometric ratios. To find the value of $\tan \dfrac{\theta }{2}$ , first we will find the value of $\cos \dfrac{\theta }{2}$ and $\sin \dfrac{\theta }{2}$ .
We will use the different formula of trigonometric identities ratio. We know that in a right angle triangle, $\tan $ is defined as the ratio of the perpendicular of the triangle to the base of the triangle. Tan is also defined as the ratio of the length of the opposite side to the length of the adjacent side. Tangent is always related to sine and cosine. $\cos \theta $ is defined in a right angle triangle as the ratio of base to hypotenuse of the right angle triangle. We will use the trigonometric identities $\cos 2\theta = 2{\cos ^2}\theta - 1$ and $\cos 2\theta = 1 - 2{\sin ^2}\theta $. The tangent function is undefined at all the points where the value of cosine is equal to zero. The domain of the tangent function is all the real numbers except $\left( {\dfrac{\pi }{2} + n\pi } \right)$ where $n$ is the integer.

Complete step by step solution:
Step: 1 to find the $\tan \dfrac{\theta }{2}$ , first we will find $\sin \dfrac{\theta }{2}$ and $\cos \dfrac{\theta }{2}$.
The given value of the function is $\cos \theta = - \dfrac{4}{5}$ .
Use the trigonometric formula $\cos 2\theta = 1 - 2{\sin ^2}\theta $ .
$\cos \theta = 1 - 2{\sin ^2}\dfrac{\theta }{2}$
Therefore,
$ \Rightarrow \dfrac{{ - 4}}{5} = 1 - 2{\sin ^2}\dfrac{\theta }{2}$
Solve the equation to find the value.
$ \Rightarrow \sin \dfrac{\theta }{2} = \pm \dfrac{3}{{\sqrt {10} }}$
Step: 2 to find the value of $\cos \dfrac{\theta }{2}$ use the trigonometric formula of $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$,
$ \Rightarrow \cos \theta = - \dfrac{4}{5}$
Therefore,
$ - \dfrac{4}{5} = 2{\cos ^2}\dfrac{\theta }{2} - 1$
Solve the equation to find the value.
$ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = 1 - \dfrac{4}{5}$
$ \Rightarrow {\cos ^2}\dfrac{\theta }{2} = \dfrac{1}{{5 \times 2}}$
$ \Rightarrow \cos \dfrac{\theta }{2} = \sqrt {\dfrac{1}{{10}}} $
Step: 3 Since $\theta $ is in the second quadrant so that $\dfrac{\theta }{2}$ will be in first quadrant. We know that both $\sin \dfrac{\theta }{2}$ and $\cos \dfrac{\theta }{2}$ are positive.
$ \Rightarrow \tan \dfrac{\theta }{2} = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}$

Substitute the value of $ \Rightarrow \cos \dfrac{\theta }{2} = \sqrt {\dfrac{1}{{10}}} $ and $\sin \dfrac{\theta }{2} = + \dfrac{3}{{\sqrt {10} }}$ in the formula to find the value of $\tan \dfrac{\theta }{2}$ .
$ \Rightarrow \tan \dfrac{\theta }{2} = \dfrac{{\dfrac{3}{{\sqrt {10} }}}}{{\dfrac{1}{{\sqrt {10} }}}}$
Simplify the equation to find the value.
$ \Rightarrow \tan \dfrac{\theta }{2} = \dfrac{3}{{\sqrt {10} }} \times \dfrac{{\sqrt {10} }}{1}$
$ \Rightarrow \tan \dfrac{\theta }{2} = 3$
Final Answer:

Therefore,
$\tan \dfrac{\theta }{2} = 3$.

Note:
Students are advised to remember that the value of tangent function in the second quadrant will be always negative. Apply the half angle formula $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$ and $\cos \theta = 1 - 2{\sin ^2}\dfrac{\theta }{2}$ very carefully to avoid any mistake. Convert the given angle into half angle so that we can apply half angle formula.