Question

How do you evaluate $\tan \left( \dfrac{3\pi }{4} \right)$?

Answer 1

We will see the definition of the tangent function. We will write the given angle as a difference of two angles. We have a formula for the expansion of tangent function of the difference of two angles. We will use this formula and then substitute the values of the terms that involve standard angles. Simplifying this expression, we will obtain the required value.

Complete step-by-step solution:
The tangent function is defined as the ratio of the sine and cosine function of an angle. This means that we have
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
We have to find the value of $\tan \left( \dfrac{3\pi }{4} \right)$. Now, we will rewrite the angle as a difference of two angles. That is, we will write $3\pi =4\pi -\pi$. Therefore, we have the following,
$\begin{aligned} & \tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \dfrac{4\pi -\pi }{4} \right) \\\ & \therefore \tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \pi -\dfrac{\pi }{4} \right)....(i) \\\ \end{aligned}$
We have a formula for expanding the tangent function of the difference of two angles. This formula is given as follows,
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
Substituting $A=\pi$ and $B=\dfrac{\pi }{4}$ in the above formula, we get
$\tan \left( \pi -\dfrac{\pi }{4} \right)=\dfrac{\tan \pi -\tan \dfrac{\pi }{4}}{1+\tan \pi \tan \dfrac{\pi }{4}}$
We know that the angles $\pi$ and $\dfrac{\pi }{4}$ are standard angles. The value of the tangent function for these two angles is $\tan \pi =0$ and $\tan \dfrac{\pi }{4}=1$. Substituting these values in the above equation, we get the following,
$\begin{aligned} & \tan \left( \pi -\dfrac{\pi }{4} \right)=\dfrac{0-1}{1+\left( 0 \right)\left( 1 \right)} \\\ & \Rightarrow \tan \left( \pi -\dfrac{\pi }{4} \right)=\dfrac{-1}{1} \\\ & \therefore \tan \left( \pi -\dfrac{\pi }{4} \right)=-1 \\\ \end{aligned}$
Substituting this value in equation $(i)$, we get
$\tan \left( \dfrac{3\pi }{4} \right)=-1$

Note: The angle $\dfrac{3\pi }{4}$ lies in the second quadrant where only the sine function is positive. So, the negative sign of the value we obtained for the tangent function is as expected. We can write the angle as $\dfrac{3\pi }{4}=\dfrac{\pi }{2}+\dfrac{\pi }{4}$, but the value for $\tan \dfrac{\pi }{2}$ is not defined because the cosine function in the denominator is zero. Therefore, we chose to write the given angle as a difference instead of a sum.