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

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Last updated date: 13th Jun 2024
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Answer
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Hint: We are given $\tan \left( \dfrac{3\pi }{4} \right)$ , we are asked to find its value, to do so we will learn about the relation between degree and the radian, then we will learn in which quadrant our coordinate will lie, once we have the quadrant will work on what sign it posses then using the value of the function in quadrant ‘I’, we will access the value of the function in that quadrant.

Complete step by step answer:
We are given $\tan \left( \dfrac{3\pi }{4} \right)$, we can see that the angle given to us is in radian.
We will change it into a degree as the degree is easy to see and understand.
Now the relation between degree and radian is given as –
180 degree $=\pi $ radian
Or
$\pi \text{ radian}=180\text{ degree}$
So, $1\text{ radian}=\dfrac{{{180}^{\circ }}}{\pi }\text{degree}$
By unitary method,
So, $\dfrac{3\pi }{4}\text{radian}=\dfrac{{{180}^{\circ }}}{\pi }\times \left( \dfrac{3\pi }{4} \right)\text{degree}$
By simplifying, we get –
$\dfrac{3\pi }{4}\text{radian}=135\text{ degree}$
Now we get the angle in degree is ${{135}^{\circ }}$ and we know that the first Quadrant is from 0 to ${{90}^{\circ }}$ .
second is from ${{90}^{\circ }}$ to ${{180}^{\circ }}$ so our angle ${{135}^{\circ }}$ will lie in the second Quadrant.
Now we will look for the sign of the tan functions in this Quadrant.
The tan function is positive in first and third Quadrant and in rest two it is negative so in Quadrant second, tan is negative.
Now we will see that ${{135}^{\circ }}$ can be written as ${{135}^{\circ }}={{180}^{\circ }}-{{45}^{\circ }}$ .
So, $\tan \left( {{135}^{\circ }} \right)=\tan \left( {{180}^{\circ }}-{{45}^{\circ }} \right)$
Now, we will use that –
$\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta $
So for $\tan \left( {{180}^{\circ }}-{{45}^{\circ }} \right)=-\tan \left( {{45}^{\circ }} \right)$
As we know that $\tan {{45}^{\circ }}=1$
So,
 $\begin{align}
  & \tan \left( {{180}^{\circ }}-{{45}^{\circ }} \right)=-\tan {{45}^{\circ }} \\
 & =-1 \\
\end{align}$
Hence, we get –
$\tan \left( {{135}^{\circ }} \right)=-1$
Or we can say that –
$\tan \left( \dfrac{3\pi }{4} \right)=-1$

Note: Remember it is easy to see this on the degree scale, but it is not necessary to always change to the degree scale, we can also solve on the radian form also.
Now $\dfrac{3\pi }{4}=\pi -\dfrac{\pi }{4}$
So, $\tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \pi -\dfrac{\pi }{4} \right)$
As we know $\tan \left( \pi -\theta \right)=-\tan \theta $
So in our case $\theta =\dfrac{\pi }{4}$
So, $\tan \left( \pi -\dfrac{\pi }{4} \right)=-\tan \left( \dfrac{\pi }{4} \right)$
As $\tan \dfrac{\pi }{4}=1$
So,
$\tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \pi -\dfrac{\pi }{4} \right)=-\tan \dfrac{\pi }{4}=-1$
Hence the value of $\tan \left( \dfrac{3\pi }{4} \right)$is -1.