Solveeit Logo
Login

Question

Question: If \({{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right)=\alpha \) and \({{\tan }^{-1}}\left( -\tan \...

If tan⁡−1(tan⁡5π4)=α{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right)=\alpha and tan⁡−1(−tan⁡2π3)=β{{\tan }^{-1}}\left( -\tan \dfrac{2\pi }{3} \right)=\beta then.
(A). α−β=7π12\alpha -\beta =\dfrac{7\pi }{12}
(B). α+β=7π12\alpha +\beta =\dfrac{7\pi }{12}
(C). 2α+3β=7π122\alpha +3\beta =\dfrac{7\pi }{12}
(D). 4α+3β=7π124\alpha +3\beta =\dfrac{7\pi }{12}

Explanation

Solution

Hint: Now you have 2 variables. So, take 2 cases. Use the concept of inverse trigonometry and find the values of α,β\alpha ,\beta . use all necessary conditions for inverse of a tangent. After getting values of α\alpha and β\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 −π2<tan⁡−1x<π2-\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:
α=tan⁡−1(tan⁡5π4)\alpha ={{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right) ………………………. (1)
β=tan⁡−1(−tan⁡(2π3))\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:
−π2<tan⁡−1x<π2-\dfrac{\pi }{2}<{{\tan }^{-1}}x<\dfrac{\pi }{2}
We need to find value of the expression given as:
tan⁡−1(tan⁡5π4){{\tan }^{-1}}\left( \tan \dfrac{5\pi }{4} \right)
By finding the value of expression, inside the bracket, we get:
tan⁡(5π4)=tan⁡(π+π4)\tan \left( \dfrac{5\pi }{4} \right)=\tan \left( \pi +\dfrac{\pi }{4} \right)
We know tan⁡(π+x)=tan⁡x\tan \left( \pi +x \right)=\tan x . So, we write
tan⁡(5π4)=tan⁡(π4)\tan \left( \dfrac{5\pi }{4} \right)=\tan \left( \dfrac{\pi }{4} \right)
So, we write tan⁡−1(tan⁡(5π4))=π4{{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{4} \right) \right)=\dfrac{\pi }{4}
By equation (1), we say that α=π4\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(−tan⁡(2π3)){{\tan }^{-1}}\left( -\tan \left( \dfrac{2\pi }{3} \right) \right) .
By this trigonometry we know tan⁡(π−x)=−tan⁡x\tan \left( \pi -x \right)=-\tan x .
So, we write tan⁡(2π3)=tan⁡(π−π3)=−tan⁡π3\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(−tan⁡2π3)=tan⁡−1(tan⁡π3)=π3{{\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 β\beta to e:
β=π3\beta =\dfrac{\pi }{3} ………………………. (4)
Now by substituting α,β\alpha ,\beta in options, we get:
Option:1 α−β=7π2\alpha -\beta =\dfrac{7\pi }{2}
π4−π3=7π12\dfrac{\pi }{4}-\dfrac{\pi }{3}=\dfrac{7\pi }{12}
By taking least common multiple on left hand side, we get:
3π−4π12=7π12\dfrac{3\pi -4\pi }{12}=\dfrac{7\pi }{12}
By simplifying the above equation, we can write it as:
−π12=7π12-\dfrac{\pi }{12}=\dfrac{7\pi }{12} (It is wrong, we can say directly)
Where, LHS≠RHSLHS\ne RHS .
So, this option is wrong.
Option-2: It is given by α+β=7π12\alpha +\beta =\dfrac{7\pi }{12} .
By substituting α,β\alpha ,\beta values into above equation, we get –
π4+π3=7π12\dfrac{\pi }{4}+\dfrac{\pi }{3}=\dfrac{7\pi }{12}
By taking least common multiple on left hand side, we get:
4π+3π12=7π12\dfrac{4\pi +3\pi }{12}=\dfrac{7\pi }{12}
By simplifying the left hand side of above equation, we get:
7π12=7π12\dfrac{7\pi }{12}=\dfrac{7\pi }{12}
So, this option is true.
Option-3: It is given by 2α+3β=7π122\alpha +3\beta =\dfrac{7\pi }{12}
By substituting values of α,β\alpha ,\beta values into above equation, we get –
2(π4)+3(π3)=7π122\left( \dfrac{\pi }{4} \right)+3\left( \dfrac{\pi }{3} \right)=\dfrac{7\pi }{12}
By simplifying the above equation, we get it as: 3π2=7π12\dfrac{3\pi }{2}=\dfrac{7\pi }{12} ,which is wrong. So, this option is wrong
Option-4: It is given by 4α+3β=7π124\alpha +3\beta =\dfrac{7\pi }{12}
By substituting values of α,β\alpha ,\beta values into above equation, we get –
4(π4)+3(π3)=7π124\left( \dfrac{\pi }{4} \right)+3\left( \dfrac{\pi }{3} \right)=\dfrac{7\pi }{12}
We can directly solve it as 2π=7π122\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 β\beta . For your convenience you can bring it out and then solve it. We convert the values 5π4,2π3\dfrac{5\pi }{4},\dfrac{2\pi }{3} as π4,π3\dfrac{\pi }{4},\dfrac{\pi }{3} in order to bring the angle in range of [−π2,π2]\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.