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 $\lambda =4$.
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