Question

Calculate without resorting to table.
$\tan 9^{\circ }-\tan {63}^{\circ }+\tan {81}^{\circ }-\tan {27}^{\circ }$

Answer 1

We know that, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\tan (A+B)(1-\tan A\tan B)=\tan A+\tan B$

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

$x=\cot 9°+\tan 9°-(\cot 27°+\tan 27°)$

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

$x=\frac{1}{\sin 9°\cos 9°}-\frac{1}{\sin 27°\cos 27°}$

$x=\frac{2}{\sin 18°}-\frac{2}{\sin 54°}=2\left[\frac{\sin 54°-\sin 18°}{\sin 54°\sin 18°}\right]$

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