Question

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

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

Answer 1

Answer: (B)

$-$1

$\cos [{\cos }^{-1}\cos \left[\frac{8\pi }{7}\right]+{\tan }^{-1}\tan \left[\frac{8\pi }{7}\right]]$ has the value equal to -
We know that the range of $co{s}^{-1}$ is $[0,\pi ]$ and that of $ta{n}^{-1}$ is $(\frac{-\pi }{2},\frac{\pi }{2})$
So, ${\cos }^{-1}\cos \left[\frac{8\pi }{7}\right]=-\frac{8\pi }{7}+2\pi$

${\tan }^{-1}\tan \left[\frac{8\pi }{7}\right]=\frac{8\pi }{7}-\pi$

Adding the two values, we get $\pi$

Thus, the question becomes $cos\pi$, which equals $-1.$