Question

Find the value of $\tan 9^{\circ }-\tan {27}^{\circ }-\tan {63}^{\circ }+\tan {81}^{\circ }.$

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (A)

4

$\tan 9°-\tan 27°-\tan 63°+\tan 81°$

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

$=\tan 9°+\tan (90°-9°)-(\tan 27°+\tan (90°-27°))$

$=\tan 9°+\cot 9°-(\tan 27°+\tan 27°)$ $[\because \cot A=\tan (90˚-A\left)\right]$

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

$=\frac{{\sin }^{2}9°+{\cos }^{2}9°}{\sin 9°\cos 9°}-\left(\frac{{\sin }^{2}27°+{\cos }^{2}27°}{\sin 27°\cos 27°}\right)$

$=\frac{1}{\sin 9°\cos 9°}-\frac{1}{\sin 27°\cos 27°}=\frac{2}{\sin 18°}-\frac{2}{\sin 54°}$ $[\because \sin 2x=2\sin x\cos x]$

$=2\left[\frac{\sin 54°-\sin 18°}{\sin 18°\sin 54°}\right]=\frac{2.2\cos 36°\sin 18°}{\sin 18°\sin (90°-36°)}$ $[\because \sin C-\sin D=2cos\frac{C+D}{2}\sin \frac{C-D}{2}]$

$=\frac{2.2\cos 36°\sin 18°}{\cos 36°\sin 18°}=4$