trigonometry - Find the value of $lambda$ in $frac{3-tan^2 {piover 7}}{1-tan^2 {piover 7}}=lambdacos{piover 7}$

Find the value of $\lambda$ in

$$\dfrac{3-\tan^2 {\pi\over 7}}{1-\tan^2 {\pi\over 7}}=\lambda\cos{\pi\over 7}$$

The numerator looks similar to expansion of $\tan 3x$, so I tried this

$$\dfrac{3\tan {\pi\over 7}-\tan^3 {\pi\over 7}}{\tan {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)}=\lambda\cos{\pi\over 7}$$

$$\dfrac{\left(3\tan {\pi\over 7}-\tan^3 {\pi\over 7}\right)\left(1-3\tan^2 {\pi\over 7}\right)}{\tan {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)\left(1-3\tan^2 {\pi\over 7}\right)}=\lambda\cos{\pi\over 7}$$

$$\dfrac{\tan {3\pi\over 7}\left(1-3\tan^2 {\pi\over 7}\right)}{\sin {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)}=\lambda$$

But I'm stuck here. Need help. Thanks in advance.

Answer

Let $a= \pi/7$ then we have

$$1+{2\over 1-\tan^2a} = \lambda \cos a$$

multiplying this with $\tan a$ we get:

$$\tan a+ \tan 2a = \lambda \sin a$$

so

$$\sin a\cos 2a +\sin 2a\cos a = \lambda \sin a \cos a \cos 2a$$

multiplying this with 4 we get

$$4\sin 3a = \lambda \sin 4a$$

Sinnce $\sin 3a = \sin 4a$ we get $\lambda =4$.