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

Last updated date: 25th Jul 2024
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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 $\tan \left[ -x \right]=-\tan x$.

We assume that ${{\tan }^{-1}}z=\alpha$. Taking the inverse form, we get $\tan \alpha =z$.
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 $\left| \alpha \right|\le \dfrac{\pi }{2}$.
The final form becomes $\tan \left( \pi -\alpha \right)=\tan \left( 2\times \dfrac{\pi }{2}-\alpha \right)=-\tan \alpha =-z$.
So, $\tan \left[ \tan \left( \pi -{{\tan }^{-1}}z \right) \right]=\tan \left[ -z \right]=-\tan z$.
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.