If $\alpha$ and $\beta$ are acute angles and $\displaystyle{\cos2\alpha=\frac{3\cos\beta-1}{3-\cos2\beta}}$, then prove that $\displaystyle{\tan\alpha=\sqrt{2}\tan\beta}$.
I tried this question by taking the formula of $\cos2\alpha$ in terms of $\tan$ (which is of degree two) but I couldn't prove it. Please suggest some hints.
Answer
Using Weierstrass substitution in either sides.
$$\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{3\cdot\dfrac{1-\tan^2\beta}{1+\tan^2\beta}-1}{3-\dfrac{1-\tan^2\beta}{1+\tan^2\beta}}$$
$$\implies\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{2-4\tan^2\beta}{2+4\tan^2\beta}$$
Using Componendo and dividendo, $$\frac{\tan^2\alpha}1=\frac{4\tan^2\beta}2$$
No comments:
Post a Comment