Question

The value of $ta{n}^{-1}\left(1\right)+ta{n}^{-1}\left(0\right)+ta{n}^{-1}\left(2\right)+ta{n}^{-1}\left(3\right)$ is equal to

• A. $\pi$
• B. $\frac{5\pi }{4}$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (A)

$ta{n}^{-1}\left(1\right)+ta{n}^{-1}\left(0\right)+ta{n}^{-1}\left(2\right)+ta{n}^{-1}\left(3\right)$
$=\frac{\pi }{4}+\pi +ta{n}^{-1}\left(\frac{2+3}{1-2\cdot 3}\right)$ (as $2\cdot 3>1$)
$=\frac{5\pi }{4}+ta{n}^{-1}\left(-1\right)$
$=\frac{5\pi }{4}-\frac{\pi }{4}=\pi$