Question

Find the value of $\tan \left[ \tan \left( \pi -{{\tan }^{-1}}z \right) \right]$?
A. $2z$
B. $-z$
C. $z$
D. None of these

Answer 1

Hint: We first assume the conversion of ${{\tan }^{-1}}z=\alpha $ which gives $\tan \alpha =z$. We use the associative angle formula to apply the associative angle. We use the identity formula of $\tan \left[ -x \right]=-\tan x$.

Complete step by step answer:
We assume that ${{\tan }^{-1}}z=\alpha $. Taking the inverse form, we get $\tan \alpha =z$.
We have two values for ratio tan to operate.
For general form of $\tan \left( x \right)$, we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha $ from that multiple of $\dfrac{\pi }{2}$ to make it equal to x.
Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha $, $k\in \mathbb{Z}$. Here we took the addition of $\alpha $. We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$.
Now we take the value of k. If it’s even then keep the ratio as tan and if it’s odd then the ratio changes to cot ratio from tan.
Then we find the position of the given angle as a quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angle lies in the first or third quadrant then the sign remains positive but if it falls in the second or fourth quadrant then the sign becomes negative.
The final form becomes $\tan \left( \pi -\alpha \right)=\tan \left( 2\times \dfrac{\pi }{2}-\alpha \right)=-\tan \alpha =-z$.
So, $\tan \left[ \tan \left( \pi -{{\tan }^{-1}}z \right) \right]=\tan \left[ -z \right]=-\tan z$.

So, the correct answer is “Option D”.

Note: There are two tan ratio operations which will give values instead of angles. The identity $\tan \left[ -x \right]=-\tan x$ works in the principal domain of ratio tan.