Question

If $ta{n^{ - 1}}\;\left( {tan \left( {\dfrac{{5\pi }}{4}} \right)} \right) = \alpha , ta{n^{ - 1}}\;\left( { - tan \left( {\dfrac{{2\pi }}{3}} \right)} \right) = \beta , $then which of the following holds.
A. $4\alpha - 4\beta = 0$
B. $4\alpha - 3\beta = 0$
C. $ \alpha > \beta $
D. None of these

Answer 1

Hint: We will evaluate the given trigonometric function to obtain the value of $\alpha $and $\beta $. Then substitute the obtained values one by one in all the given options to determine which option holds the values of the given trigonometric function.

Formula Used:
The property of inverse trigonometric function used $ta{n^{ - 1}}\;\left( {tan x} \right) = x$
The value of the trigonometric function is obtained in the first quadrant by adding or subtracting the angle by π. As shown below
${\alpha = ta{n^{ - 1}}\;\left( {tan \left( {\dfrac{{5\pi }}{4}\;} \right)} \right)}$
$\alpha \:\:=\:\:tan^{\:-\:1}\left(tan\:\left(\:\pi \:\:\:+\:\:\left(\:\dfrac{\pi \:}{4}\:\right)\:\right)\right)$
$\alpha \:\:=\:\:tan^{\:-\:1}\left(\:tan\dfrac{\pi \:}{4}\:\right)$

Complete step by step solution:
The given trigonometric function:
$\alpha = ta{n^{ - 1}}\;\left( {tan \left( {\dfrac{{5\pi }}{4}\;} \right)} \right)$
Evaluate the trigonometric angle to the first quadrant:
$\alpha = ta{n^{ - 1}}\;(tan \left( {\pi + \left( {\dfrac{\pi }{4}} \right)} \right)$
Evaluating the angle further, we get
$\alpha = ta{n^{ - 1}}\;\left( {tan\dfrac{\pi }{4}} \right)$
So, the value of $\alpha $will be:
$\alpha = \dfrac{\pi }{4}$
Now, on similarly determine the value of $\beta $
The given trigonometric function:
$\beta = ta{n^{ - 1}}\;\left( { - tan \left( {\dfrac{{2\pi }}{3}} \right)} \right)$
Evaluate the trigonometric angle to the first quadrant:
$\beta = ta{n^{ - 1}}\;( - tan \left( {\pi - \left( {\dfrac{\pi }{3}} \right)} \right)$
Evaluating the angle further, we get
$\beta = ta{n^{ - 1}}\;\left( {tan \left( {\dfrac{\pi }{3}} \right)} \right)$
So, the value of $\beta $will be:
$\beta = \left( {\dfrac{\pi }{3}} \right)$
Since the value of $\alpha $and $\beta $ are known, we will substitute the values in the option to determine which of the options holds good for given trigonometric function.

Substitute the values of $\alpha $and $\beta $ in option A that is $4\alpha - 4\beta = 0$, we get
$4\left( {\dfrac{\pi }{4}} \right) -{{ 4}}\left( {\dfrac{\pi }{3}} \right) \ne 0$
So, the option A does not hold for the given trigonometric function.

Now substitute the values of $\alpha $ and $\beta $ in option B that is $4\alpha - 3\beta = 0$, we get, $4\left( {\dfrac{\pi }{4}} \right) - 3\left( {\dfrac{\pi }{3}} \right) = 0$
So, option B does not hold for the given trigonometric function.

Now substitute the values of $\alpha $ and $\beta $ in option C that is $ \alpha > \beta $, we get
$\left( {\dfrac{\pi }{4}} \right){{ < }}\left( {\dfrac{\pi }{3}} \right) $
So, the option C does not hold for the given trigonometric function.

Option ‘B’ is correct

Note: One can make a mistake in substituting the values of $\alpha $ and $\beta $. This problem can also be solved by directly putting the trigonometric function in the option and then evaluating it.