Trig2.12

If $\tan \beta=\frac{n\tan\alpha}{1+(1-n)\tan^2 \alpha}\,$ then show that $\tan (\alpha-\beta)=(1-n)\tan \alpha\,$

Given $\tan \beta=\frac{n\tan\alpha}{1+(1-n)\tan^2 \alpha}\,$

$\tan (\alpha-\beta)=\frac{\tan \alpha+\tan \beta}{1+\tan \alpha \tan \beta}\,$

$\tan (\alpha-\beta)=\frac{\tan \alpha-\frac{n\tan \alpha}{1+(n-1)\tan^2 \alpha}}{1+\tan \alpha \frac{n\tan \alpha}{1+(n-1)\tan^2 \alpha}}\,$

$\frac{\tan \alpha+(1-n)\tan^3 \alpha-n\tan \alpha}{1+\tan^2 \alpha-n\tan^2 \alpha+n\tan^2 \alpha}\,$

$\frac{[1+(1-n)\tan^2 \alpha-n]\tan \alpha}{1+\tan^2 \alpha}\,$

$\frac{(1-n)(1+\tan^2 \alpha)\tan \alpha}{1+\tan^2 \alpha}=(1-n)\tan \alpha\,$

Therefore $\tan (\alpha-\beta)=(1-n)\tan \alpha\,$