If $\alpha= \frac{\pi}{7}$ and also if $\frac{3- \tan^2 (\alpha)}{1 - \tan^2 (\alpha)} = \lambda \cos(\alpha)$. Then find $\lambda$. Given that
$$\frac{3- \tan^2 \left(\dfrac{\pi}{7}\right)}{1 - \tan^2 \left(\dfrac{\pi}{7}\right)} = \lambda \cos\left(\dfrac{\pi}{7}\right)$$
Then find the value of $\lambda$.
 A: Notice, setting the value of $\alpha=\pi/7$, one can simplify LHS as follows   $$LHS=\frac{3-\tan^2(\alpha)}{1-\tan^2(\alpha)}$$
$$=\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}$$
$$=\frac{1-\tan^2\frac{\pi}{7}+2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2\tan\frac{\pi}{7}}{\left(1-\tan^2\frac{\pi}{7}\right)\tan\frac{\pi}{7}}$$
$$=1+\left(\frac{2\tan\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}\right)\frac{1}{\tan\frac{\pi}{7}}$$
using $\color{blue}{\frac{2\tan A}{1-\tan^2A}=\tan2A}$, 
$$=1+\frac{\tan\frac{2\pi}{7}}{\tan\frac{\pi}{7}}$$
$$=1+\frac{\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
using $\color{blue}{\sin A\cos B+\sin B\cos A=\sin(A+B)}$, 
$$=\frac{\sin\frac{3\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\left(\pi-\frac{3\pi}{7}\right)}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\frac{4\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{2\sin\frac{2\pi}{7}}{\sin\frac{\pi}{7}}$$
$$=\frac{2\cdot 2\sin\frac{\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}}$$
$$=4\cos \frac{\pi}{7}$$
now, $RHS=\lambda\cos(\alpha)=\lambda\cos\frac{\pi}{7}$,
Now, equating LHS & RHS, we have $$4\cos\frac{\pi}{7}=\lambda\cos\frac{\pi}{7}$$ 
 $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\lambda=4}}$$
A: Simply divide both sides of the equation by $\cos(\pi/7)$
$\lambda = \frac{3-\tan^2(\pi/7)}{\cos(\pi/7)(1-\tan^2(\pi/7)}$ = 3.999...
