Question

tan ${81}^0$ - tan ${63}^0$ - tan ${27}^0$ + tan $9^0$ equals.

• A. 6
• B. 0
• C. 2
• D. 4

Answer 1

Answer: (D)

4

$\tan 81-\tan 63-\tan 27+\tan 9$

$\tan 81+\tan 9-(\tan 63+\tan 27)$

$\cot 9+\tan 9-(\cot 27+\tan 27)\to \tan A=\cot (90-A)0^o\le A\le {90}^o$

$\frac{\cos 9}{\sin 9}+\frac{\sin 9}{\cos 9}-(\frac{\cos 27}{\sin 27}+\frac{\sin 27}{\cos 27})$

$\frac{1}{\sin 9\cos 9}-\frac{1}{\sin 27\cos 27}$

$\frac{2}{\sin 18}-\frac{2}{\sin 54}\to \sin 2A=2\sin A\cos A$

General form, $\sin 18=\frac{\sqrt{5}-1}{4}\sin \sqrt{4}=\frac{\sqrt{5}+1}{4}$

$\therefore \frac{2}{\sqrt{5}-1}\times 4-\frac{2}{\sqrt{5}+1}\times 4$

$\frac{8}{\sqrt{5}-1}-\frac{8}{\sqrt{5}+1}=8\times \frac{2}{4}=4$

Solution to $\sin {18}^o$

$\theta ={18}^o\therefore 5\theta ={90}^{o}2\theta +3\theta ={90}^{o}2\theta =90-3\theta$

$\therefore \sin 2\theta =\cos 3\theta$

$\therefore 2\sin \theta \cos \theta =4{\theta }^3\theta -3\cos \theta$

$\cos \theta (2\sin \theta -4{\cos }^2\theta +3)=0$

$2\sin \theta -4(1-{\sin }^2\theta )+3=0$

$2\sin \theta -4(1-{\sin }^2\theta )+3=0$

$4{\sin }^2\theta +2\sin \theta -1=0$

$\therefore \sin \theta =\frac{-2\pm \sqrt{4+16}}{8}=\pm \frac{\sqrt{5}-1}{4}$ But $\sin {18}^o>0$

$\therefore \sin \theta =\sin {18}^o=\frac{\sqrt{5}-1}{4}$