Question

The value of ${\tan }^{-1}\left(\frac{\cos \left(\frac{15\pi }{4}\right)-1}{\sin \left(\frac{\pi }{4}\right)}\right)$ is equal to

• A. $-\frac{\pi }{4}$
• B. $-\frac{\pi }{8}$
• C. $-\frac{5\pi }{12}$
• D. $-\frac{4\pi }{9}$

Answer 1

Answer: (B)

${\tan }^{-1}\left[\frac{\cos \left(4\pi -\frac{\pi }{4}\right)-1}{\sin \frac{\pi }{4}}\right]$
$\Rightarrow {\tan }^{-1}\left(\frac{\cos \frac{\pi }{4}-1}{\sin \frac{\pi }{4}}\right)$
${\tan }^{-1}\left(\frac{1-\sqrt{2}}{1}\right)=-\frac{\pi }{8}$