Question

$co{s}^{-1}\left(\frac{-1}{2}\right)-2si{n}^{-1}\left(\frac{1}{2}\right)+3co{s}^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4ta{n}^{-1}\left(-1\right)$ equals

• A. $\frac{19\pi }{12}$
• B. $\frac{35\pi }{12}$
• C. $\frac{47\pi }{12}$
• D. $\frac{43\pi }{12}$

Answer 1

Answer: (D)

${\cos }^{-1}\left(-\frac{1}{2}\right)-2{\sin }^{-1}\left(\frac{1}{2}\right)+3{\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)-4{\tan }^{-1}(-1)$
$=\pi -{\cos }^{-1}\left(\frac{1}{2}\right)-2\left(\frac{\pi }{6}\right)+3\left(\pi -{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)+4{\tan }^{-1}(1)$
$=\pi -\frac{\pi }{3}-\frac{\pi }{3}+3\left(\pi -\frac{\pi }{4}\right)+4.\frac{\pi }{4}$
$=\frac{\pi }{3}+3\pi -\frac{3\pi }{4}+\pi$
$=\frac{43\pi }{12}$