Question

The value of $\frac{\cos 15\pi +\tan \left(\frac{5\pi }{4}\right)}{\sin 16\pi +\sec 6\pi }$ is

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $-1$

Answer 1

The correct option is A $0$

$\frac{\cos 15\pi +\tan \left(\frac{5\pi }{4}\right)}{\sin 16\pi +\sec 6\pi }$

$=\frac{\cos (14\pi +\pi )+\tan (\pi +\frac{\pi }{4})}{\sin (16\pi +0)+\sec (6\pi +0)}$

$=\frac{\cos \pi +\tan \left(\frac{\pi }{4}\right)}{\sin 0+\sec 0}$

$=\frac{-1+1}{0+1}=0$