[Math] Given $\tan\beta=\frac{n\sin\alpha\cos\alpha}{1-n\sin^2\alpha}$, show that $\tan(\alpha-\beta)=(1-n)\tan\alpha$

trigonometry

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

Best Answer

As we need to eliminate $\beta,$ write $\tan\beta=\tan\{\alpha-(\alpha-\beta)\}$ and expand.

For the RHS, $$\frac{n\sin\alpha\cos\alpha}{1-n\sin^2\alpha}=\dfrac{\dfrac{n\sin\alpha\cos\alpha}{\cos^2\alpha}}{\dfrac{1-n\sin^2\alpha}{\cos^2\alpha}}=\dfrac{n\tan^2\alpha}{1+(1-n)\tan^2\alpha}$$

Related Question