Question

If ${{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right)=\alpha $ and ${{\tan }^{-1}}\left( -\tan \dfrac{2\pi }{3} \right)=\beta $ then.
(A). $\alpha -\beta =\dfrac{7\pi }{12}$
(B). $\alpha +\beta =\dfrac{7\pi }{12}$
(C). $2\alpha +3\beta =\dfrac{7\pi }{12}$
(D). $4\alpha +3\beta =\dfrac{7\pi }{12}$

Answer 1

Hint: Now you have 2 variables. So, take 2 cases. Use the concept of inverse trigonometry and find the values of $\alpha ,\beta $ . use all necessary conditions for inverse of a tangent. After getting values of $\alpha $ and $\beta $ substitute them in the options one by one and check which of them is/are true. The statements which are true will be our result. Use the condition of $-\dfrac{\pi }{2}<{{\tan }^{-1}}x<\dfrac{\pi }{2}$ .

Complete step-by-step answer:

Given 2 variables in the question, can be written as follows:
$\alpha ={{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right)$ ………………………. (1)
\[\beta ={{\tan }^{-1}}\left( -\tan \left( \dfrac{2\pi }{3} \right) \right)\] ……………….. (2)
Case-1: We will solve for equation (1) in this case:
The main point used from inverse trigonometry here is:
$-\dfrac{\pi }{2}<{{\tan }^{-1}}x<\dfrac{\pi }{2}$
We need to find value of the expression given as:
${{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right)$
By finding the value of expression, inside the bracket, we get:
$\tan \left( \dfrac{5\pi }{4} \right)=\tan \left( \pi +\dfrac{\pi }{4} \right)$
We know $\tan \left( \pi +x \right)=\tan x$ . So, we write
$\tan \left( \dfrac{5\pi }{4} \right)=\tan \left( \dfrac{\pi }{4} \right)$
So, we write ${{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{4} \right) \right)=\dfrac{\pi }{4}$
By equation (1), we say that $\alpha =\dfrac{\pi }{4}$ ……………… (3)
Case-2: We will solve for equation (2) in this are we need the value of expression given by: \[{{\tan }^{-1}}\left( -\tan \left( \dfrac{2\pi }{3} \right) \right)\] .
By this trigonometry we know $\tan \left( \pi -x \right)=-\tan x$ .
So, we write $\tan \left( \dfrac{2\pi }{3} \right)=\tan \left( \pi -\dfrac{\pi }{3} \right)=-\tan \dfrac{\pi }{3}$
By substituting this in our equation, we get it as:
${{\tan }^{-1}}\left( -\tan \dfrac{2\pi }{3} \right)={{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}$
By equation (2), we can say value of $\beta $ to e:
$\beta =\dfrac{\pi }{3}$ ………………………. (4)
Now by substituting $\alpha ,\beta $ in options, we get:
Option:1 $\alpha -\beta =\dfrac{7\pi }{2}$
\[\dfrac{\pi }{4}-\dfrac{\pi }{3}=\dfrac{7\pi }{12}\]
By taking least common multiple on left hand side, we get:
$\dfrac{3\pi -4\pi }{12}=\dfrac{7\pi }{12}$
By simplifying the above equation, we can write it as:
$-\dfrac{\pi }{12}=\dfrac{7\pi }{12}$ (It is wrong, we can say directly)
Where, $LHS\ne RHS$ .
So, this option is wrong.
Option-2: It is given by $\alpha +\beta =\dfrac{7\pi }{12}$ .
By substituting $\alpha ,\beta $ values into above equation, we get –
\[\dfrac{\pi }{4}+\dfrac{\pi }{3}=\dfrac{7\pi }{12}\]
By taking least common multiple on left hand side, we get:
$\dfrac{4\pi +3\pi }{12}=\dfrac{7\pi }{12}$
By simplifying the left hand side of above equation, we get:
$\dfrac{7\pi }{12}=\dfrac{7\pi }{12}$
So, this option is true.
Option-3: It is given by $2\alpha +3\beta =\dfrac{7\pi }{12}$
By substituting values of $\alpha ,\beta $ values into above equation, we get –
\[2\left( \dfrac{\pi }{4} \right)+3\left( \dfrac{\pi }{3} \right)=\dfrac{7\pi }{12}\]
By simplifying the above equation, we get it as: \[\dfrac{3\pi }{2}=\dfrac{7\pi }{12}\] ,which is wrong. So, this option is wrong
Option-4: It is given by $4\alpha +3\beta =\dfrac{7\pi }{12}$
By substituting values of $\alpha ,\beta $ values into above equation, we get –
\[4\left( \dfrac{\pi }{4} \right)+3\left( \dfrac{\pi }{3} \right)=\dfrac{7\pi }{12}\]
We can directly solve it as $2\pi =\dfrac{7\pi }{12}$ , which is wrong. So, this option is wrong.
Therefore, option (b) is the correct answer for the given equation.

Note: Generally students confuse the “-“ sign inside the $\beta $ . For your convenience you can bring it out and then solve it. We convert the values $\dfrac{5\pi }{4},\dfrac{2\pi }{3}$ as $\dfrac{\pi }{4},\dfrac{\pi }{3}$ in order to bring the angle in range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ . This point is very important. While checking options, be careful at every point because this determines our result.