Question

$\cot \frac{\pi }{32}-\tan \frac{\pi }{32}-2\tan \frac{\pi }{16}-4\tan \frac{\pi }{8}=$

• A. $4$
• B. $8$
• C. $2$
• D. $6$

Answer 1

The correct option is B $8$

$\cot \frac{\pi }{32}-\tan \frac{\pi }{32}-2\tan \frac{\pi }{16}-4\tan \frac{\pi }{8}$

$=\frac{2({\cos }^2\frac{\pi }{32}-{\sin }^2\frac{\pi }{32})}{2\sin \frac{\pi }{32}\cos \frac{\pi }{32}}-2\tan \frac{\pi }{16}-4\tan \frac{\pi }{8}$

$=2\cot \frac{\pi }{16}-2\tan \frac{\pi }{16}-4\tan \frac{\pi }{8}$

$=2\times 2\left[\frac{{\cos }^2\frac{\pi }{16}-{\sin }^2\frac{\pi }{16}}{2\cos \frac{\pi }{16}\sin \frac{\pi }{16}}\right]-4\tan \frac{\pi }{8}$

$=4\cot \frac{\pi }{8}-4\tan \frac{\pi }{8}$

$=8\times \left[\frac{{\cos }^2\frac{\pi }{8}-{\sin }^2\frac{\pi }{8}}{2\sin \frac{\pi }{8}\cos \frac{\pi }{8}}\right]$

$=8\left[\cot \frac{\pi }{4}\right]$
$=8$