# Can $\frac{2u\sin(\alpha-\beta)}{g\cos\beta}=\frac{u\cos(\alpha-\beta)}{g\sin\beta}$ be reduced to $2\tan(\alpha-\beta)\tan\beta=1$?

Can $\frac{2u\sin(\alpha-\beta)}{g\cos\beta}=\frac{u\cos(\alpha-\beta)}{g\sin\beta}$ be reduced to $2\tan(\alpha-\beta)\tan\beta=1$?

The closest I can get is: $$\tan\alpha\tan\beta-2\tan^2{\beta}=1$$

Is this correct so far?

If not where did I go wrong? If so where does one go from here?

• Just multiply both sides by the reciprocal of the RHS. – Chappers Jul 14 '17 at 9:54
• Great , I see that now. But can my final expression also be reduced to the correct answer? – Kantura Jul 14 '17 at 10:07

$2(\tan({\alpha-\beta}))\tan\beta=1$

$⇔2(\tan\alpha-\tan\beta)\tan\beta=1+\tan\alpha\tan\beta$

$⇔\tan\alpha\tan\beta-2\tan^2\beta=1$

Cancelling $$u,g\ne 0$$ we have $$2\tan(\alpha-\beta)\tan(\beta)=1$$ (after dividing by $\cos(\alpha-\beta)$ and multiplying by $\sin(\beta)$).

• Great , I see that now. But can my final expression also be reduced to the correct answer? – Kantura Jul 14 '17 at 10:06