Question

is tanα=3tanβ, then prove that tan (α-β)=sin2β/5-cos2β

Answer 1

We know that $\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta}{1+\tan \alpha \tan \beta}$. Given $\tan \alpha =3 \tan \beta$. Now, plugging this equation in the formula,

$\tan (\alpha -\beta )=\frac{3 \tan \beta -\tan \beta}{1+3 \tan \beta \tan \beta} \\ \tan (\alpha -\beta )=\frac{2 \tan \beta }{1+3 \tan ^2\beta } \\ \tan (\alpha -\beta )=\frac{2 \tan \beta \cos^2 \beta }{\cos^2 \beta+3 \tan ^2\beta \cos^2 \beta} \\ \tan (\alpha -\beta )=\frac{2 \sin \beta \cos \beta }{\cos^2 \beta+3 \sin^2 \beta} \\ \tan (\alpha -\beta )=\frac{2 \sin \beta \cos \beta }{1+2 \sin^2 \beta}$

$\tan (\alpha -\beta )=\frac{ \sin2 \beta }{2+2 \sin^2 \beta-1} \\ \tan (\alpha -\beta )=\frac{ \sin2 \beta }{2-(1-2 \sin^2 \beta)} \\ \tan (\alpha -\beta )=\frac{ \sin2 \beta }{2-\cos 2\beta}$