Question

Find the value of $\tan \dfrac{{3\pi }}{4}$.

Answer 1

Hint: Angle $\dfrac{{3\pi }}{4}$ is an obtuse angle and is located in the 2nd quadrant. Use a trigonometric quadrant rule to find the value.

Complete step by step solution:
We can write $\tan \dfrac{{3\pi }}{4}$ = $\tan \dfrac{{\left( {4 - 1} \right)\pi }}{4}$
=> $\tan \dfrac{{3\pi }}{4}$= $\tan \dfrac{{\left( {4\pi - \pi } \right)}}{4}$
=> $\tan \dfrac{{3\pi }}{4}$= $\tan \left( {\dfrac{{4\pi }}{4} - \dfrac{\pi }{4}} \right)$
Now in $\tan \left( {\dfrac{{4\pi }}{4} - \dfrac{\pi }{4}} \right)$ we can cancel 4 from numerator and denominator from ${\sin ^{ - 1}}\theta $
We get, $\tan \dfrac{{3\pi }}{4}$= $\tan \left( {\pi - \dfrac{\pi }{4}} \right)$ ……equation (1)
As we know from trigonometric identity
$\tan \left( {\pi - x} \right)$ = -$\tan x$
So we can write $\tan \left( {\pi - \dfrac{\pi }{4}} \right)$ = -$\tan \dfrac{\pi }{4}$……equation (2)
So putting equation (2) in equation (1) we get
$\tan \dfrac{{3\pi }}{4}$ = -$\tan \dfrac{\pi }{4}$
And we know, $\tan \dfrac{\pi }{4}$= 1

So $\tan \dfrac{{3\pi }}{4}$= -1

Note: Proof for $\tan \left( {\pi - x} \right)$
We know $\tan x$= $\dfrac{{\sin x}}{{\cos x}}$
So, $\tan \left( {\pi - x} \right)$ = $\dfrac{{\sin (\pi - x)}}{{\cos (\pi - x)}}$
And we know, $\sin \left( {\pi - x} \right)$ = $\sin x$
And \[\cos \left( {\pi - x} \right)\] = -$\cos x$
So, $\tan \left( {\pi - x} \right)$ = $\dfrac{{\sin x}}{{ - \cos x}}$
$\tan \left( {\pi - x} \right)$ = - $\tan x$.