Question

$\cos \left({\cos }^{-1}\cos \left(\frac{8\pi }{7}\right)+{\tan }^{-1}\tan \left(\frac{8\pi }{7}\right)\right)$ has the value equal to

• A. 1
• B. -1
• C. $\cos \frac{\pi }{7}$
• D. 0

Answer 1

Answer: (B)

${\cos }^{-1}\left(\cos \frac{8\pi }{7}\right)={\cos }^{-1}\left(\cos \left(2\pi -\frac{8\pi }{7}\right)\right)={\cos }^{-1}\left(\cos \frac{6\pi }{7}\right)=\frac{6\pi }{7}$
${\tan }^{-1}\left(\tan \frac{8\pi }{7}\right)={\tan }^{-1}\left(\tan \frac{\pi }{7}\right)=\frac{\pi }{7}$