Question

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

• A. $(1+n)\tan \alpha$
• B. $(1-n)\tan \alpha$
• C. $-(1+n)\tan \alpha$
• D. $-(1-n)\tan \alpha$

Answer 1

The correct option is B $(1-n)\tan \alpha$

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

Putting value of $\tan \beta$ from given condition

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

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

$\Rightarrow \tan (\alpha -\beta )=(1-n)\tan \alpha$

Hence option $B$ is correct.