Question

If ${\tan }^{-1}\left(\tan \left(\frac{5\pi }{4}\right)\right)=\alpha$ and ${\tan }^{-1}\left(-\tan \left(\frac{2\pi }{3}\right)\right)=\beta$ then

• A. $4\alpha -4\beta =0$
• B. $4\alpha -3\beta =0$
• C. $\alpha >\beta$
• D. None of these

Answer 1

The correct option is B

$4\alpha -3\beta =0$

Explanation for the correct option:

Step 1: Calculate the value of $\alpha$

Let, $\begin{array}{c}{\tan }^{-1}\left(\tan x\right)=x\end{array}$

$\begin{array}{c}\therefore \alpha ={\tan }^{-1}\left(\tan \left(\frac{5\pi }{4}\right)\right)\\ ={\tan }^{-1}\left(\tan \left(\pi +\frac{\pi }{4}\right)\right)\\ ={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}\right)\right)\\ =\frac{\pi }{4}\end{array}$

$\therefore 4\alpha =\pi \mathrm{...}\left(1\right)$

Step 2 : Calculate the value of $\beta$

$\begin{array}{c}\because \beta ={\tan }^{-1}\left(-\tan \left(\frac{2\pi }{3}\right)\right)\\ ={\tan }^{-1}\left(-\tan \left(\pi -\frac{\pi }{3}\right)\right)\\ ={\tan }^{-1}\left(\tan \left(\frac{\pi }{3}\right)\right)\\ =\frac{\pi }{3}\end{array}$

$\therefore 3\beta =\pi \mathrm{...}\left(2\right)$

Step 3: Equating the equation (1) and (2)

$\begin{array}{l}\therefore \pi =\pi \\ \Rightarrow 4\alpha =3\beta \end{array}$

Hence option (B) is correct