Question

$\tan 9^o-\tan {27}^o-\tan {63}^o+\tan {81}^o=$

• A. 1/2
• B. 2
• C. 4
• D. 8

Answer 1

The correct option is C 4

$\tan 9^o-\tan {27}^o-\tan {63}^o+\tan {81}^o=\tan 9^o+\tan {81}^o-(\tan {27}^o+\tan {63}^o)$
$=\frac{\sin 9}{\cos 9}+\frac{\sin 81}{\cos 81}-(\frac{\sin 27}{\cos 27}+\frac{\sin 63}{\cos 63})$
$=\frac{1}{\cos 9\cos 81}-\left(\frac{1}{\cos 27\cos 63}\right)$
$=\frac{2}{\cos 90+\cos 72}-\frac{2}{\cos 90+\cos 36}$
$=\frac{2\times 4}{\sqrt{5}-1}-\frac{2\times 4}{\sqrt{5}+1}=4$