Question

Prove $\tan 9°-\tan 27°-\tan 63°+\tan 81°=4$

Answer 1

Determine the proving$\tan 9°-\tan 27°-\tan 63°+\tan 81°=4$

Solve the L.H.S part:

$\begin{array}{rcl}& & \tan 9°–\tan 27°–\tan 63°+\tan 81°\\ & =& \tan 81+\tan 9–(\tan 63+\tan 27)\\ & =& cot9+\tan 9–(cot27+\tan 27)\left[\therefore \tan (90–x)=\cot x\right]\\ & =& \left(\frac{\cos 9}{\sin 9}\right)+\left(\frac{\sin 9}{\cos 9}\right)–\left[\right(\frac{\cos 27}{\sin 27})+(\frac{\sin 27}{\cos 27}\left)\right]\left[\because \cot A=\frac{\cos A}{\sin A}\mathrm{and}\tan A=\frac{\sin A}{\cos A}\right]\\ & =& \frac{({\cos }^{2}9+{\sin }^{2}9)}{\sin 9\cos 9}–\left[\frac{({\cos }^{2}27+{\sin }^{2}27)}{\sin 27\cos 27}\right]\mathbf{}\left[\because {\sin }^{2}A+{\cos }^{2}A=1\right]\\ & =& \frac{1}{\sin 9\cos 9}–\frac{1}{\sin 27\cos 27}\left[\because \sin 2A=2\sin \mathrm{Acos}A\right]\\ & =& \frac{2}{\sin 18}–\frac{2}{\sin 54}\left[\because \sin 2\left(9\right)=\sin 18\mathrm{and}\sin 2\left(27\right)=\sin 54\right]\\ & =& \frac{2(\sin 54–\sin 18)}{\sin 54.\sin 18}\mathbf{}\left[\because \sin C–\sin D=2\cos \frac{C+D}{2}.\sin \frac{(C–D)}{2}\right]\\ & =& \frac{2(2\cos 36.\sin 18)}{\cos 36.\sin 18}\\ & =& \frac{4\cos 36.\sin 18}{\cos 36.\sin 18}\\ & =& 4\end{array}$

Hence, the L.H.S= R.H.S.