Question

Calculate the value $\tan \left(9^0\right)-\tan \left({27}^0\right)-\tan \left({63}^0\right)+\tan \left({81}^0\right)$

Answer 1

$\tan \left(9^{\circ }\right)-\tan \left({27}^{\circ }\right)-\tan \left({63}^{\circ }\right)+\tan \left({81}^{\circ }\right)$

$=\tan \left(9^{\circ }\right)+\tan \left({81}^{\circ }\right)-\tan \left({27}^{\circ }\right)-\tan \left({63}^{\circ }\right)$

$=\tan \left(9^{\circ }\right)+\tan \left({90}^{\circ }-9^{\circ }\right)-\tan \left({27}^{\circ }\right)-\tan \left({90}^{\circ }-{27}^{\circ }\right)$

$=\tan \left(9^{\circ }\right)+\cot \left(9^{\circ }\right)-\left[\tan \left({27}^{\circ }\right)+\cot \left({27}^{\circ }\right)\right]-----\left(1\right)$

We can express it as

$\begin{array}{rcl}\tan \left(9^{\circ }\right)+\cot \left(9^{\circ }\right)& =& \frac{1}{\sin \left(9^{\circ }\right)\cos \left(9^{\circ }\right)}\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta }\text{and cot}\theta =\frac{\cos \theta }{\sin \theta }\right]\\ & =& \frac{2}{\sin \left({18}^{\circ }\right)}-----\left(2\right)\left[\because \sin 2\theta =2\sin \theta \cos \theta \right]\\ \tan \left({27}^{\circ }\right)+\cot \left({27}^{\circ }\right)& =& \frac{1}{\sin \left({27}^{\circ }\right)\cos \left({27}^{\circ }\right)}\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta }\text{and cot}\theta =\frac{\cos \theta }{\sin \theta }\right]\\ & =& \frac{2}{\sin \left({54}^{\circ }\right)}\left[\because \sin 2\theta =2\sin \theta \cos \theta \right]\\ & =& \frac{2}{\cos \left({36}^{\circ }\right)}------\left(3\right)\left[\because \sin \left({90}^{°}-\theta \right)=\cos \theta \right]\\ & & \end{array}$

Substitute $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$, we get

$\begin{array}{rcl}\tan \left(9^{\circ }\right)+\cot \left(9^{\circ }\right)-\left[\tan \left({27}^{\circ }\right)+\cot \left({27}^{\circ }\right)\right]& =& \frac{2}{\sin \left({18}^{\circ }\right)}-\frac{2}{\cos \left({36}^{\circ }\right)}\\ & =& 4\left[\because \sin \left({18}^{\circ }\right)=\frac{\sqrt{5}-1}{4}\text{and}\cos \left({36}^{\circ }\right)\text{=}\frac{\sqrt{5}+1}{4}\right]\end{array}$

Therefore, the value of $\tan \left(9^{\circ }\right)-\tan \left({27}^{\circ }\right)-\tan \left({63}^{\circ }\right)+\tan \left({81}^{\circ }\right)=4$.