$ an 81^{\circ}- an 63^{\circ}- an 27^{\circ}+ an 9^{\circ}$ equals to

Question

$	an 81^{\circ}-	an 63^{\circ}-	an 27^{\circ}+	an 9^{\circ}$ equals to

Aicharya · Accepted Answer

$\begin{aligned} & 	an 81^{\circ}-	an 63^{\circ}-	an 27^{\circ}+	an 9^{\circ} \ & =\left(	an 81^{\circ}+	an 9^{\circ}ight)-\left[	an \left(63^{\circ}ight)+	an \left(27^{\circ}ight)ight] \ & =\left[\frac{\sin 81^{\circ}}{\cos 81^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}ight]-\left[\frac{\sin 63^{\circ}}{\cos 63^{\circ}}+\frac{\sin 27^{\circ}}{\cos 27^{\circ}}ight] \ & =\left[\frac{\sin 81^{\circ} \cos 9^{\circ}+\sin 9^{\circ} \cos 81^{\circ}}{\cos 9^{\circ} \cos 81^{\circ}}ight] \ & -\left[\frac{\sin 63^{\circ} \cos 27^{\circ}+\sin 27^{\circ} \cos 63^{\circ}}{\cos 27^{\circ} \cos 63^{\circ}}ight] \ & =\frac{\left[\sin \left(81^{\circ}+9^{\circ}ight)ight]}{\cos 9^{\circ} \cos \left(90^{\circ}-90^{\circ}ight)} \ & =\frac{\left[\sin \left(63^{\circ}+27^{\circ}ight)ight]}{\cos 27^{\circ} \cos \left(90^{\circ}-27^{\circ}ight)} \ & =\frac{1}{\sin 9^{\circ} \cos 9^{\circ}} 	imes \frac{2}{2}-\frac{1}{\sin 27^{\circ} \cos 27^{\circ}} 	imes \frac{2}{2} \ & =\frac{2}{\cos 27^{\circ} \sin 27^{\circ}}-\frac{2}{\sin 54^{\circ}}\end{aligned}$ $\begin{aligned} & =2\left[\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}ight] \ & =2\left[\frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}}ight] \ & =4\end{aligned}$

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$ equals to

Solution