Given $$\tan\beta=\frac{n\sin\alpha\cos\alpha}{1-n\sin^2\alpha},$$ show that $\tan(\alpha-\beta)=(1-n)\tan\alpha$.
Now I used formula for $\tan(A-B)$ and then put value of $\tan\beta$ and so I am able to prove the answer, but I asked this question because I wanted to know if there are other ways to do it.
thanks
With $$\tan\beta=\frac{n\sin\alpha\cos\alpha}{1-n\sin^2\alpha}$$ we write $$n=\frac{\tan\beta}{\sin\alpha\cos\alpha+\tan\beta\sin^2\alpha}$$ so \begin{eqnarray} \color{blue}{1-n}&=& \frac{\sin\alpha\cos\alpha+\tan\beta\sin^2\alpha-\tan\beta}{\sin\alpha\cos\alpha+\tan\beta\sin^2\alpha}\\ &=& \frac{\sin\alpha\cos\alpha-\cos^2\alpha\tan\beta}{\sin\alpha\cos\alpha+\tan\beta\sin^2\alpha}\\ &=& \frac{\tan\alpha-\tan\beta}{\tan\alpha+\tan\beta\tan^2\alpha}\\ &=& \color{blue}{\frac{1}{\tan\alpha}\tan(\alpha-\beta)} \end{eqnarray}
As we need to eliminate $\beta,$ write $\tan\beta=\tan\{\alpha-(\alpha-\beta)\}$ and expand.
For the RHS, $$\frac{n\sin\alpha\cos\alpha}{1-n\sin^2\alpha}=\dfrac{\dfrac{n\sin\alpha\cos\alpha}{\cos^2\alpha}}{\dfrac{1-n\sin^2\alpha}{\cos^2\alpha}}=\dfrac{n\tan^2\alpha}{1+(1-n)\tan^2\alpha}$$
Hint...you can rewrite $\tan\beta$ as $$\frac{n}{1+(1-n)\tan\alpha}$$ then use the compound angle formula for $\tan(\alpha-\beta)$