Question

If $tan\beta =\frac{nsin\alpha cos\alpha }{1-nco{s}^2\alpha }$ then, $\tan (\alpha +\beta )$ is equal to

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

Answer 1

The correct option is D $\frac{-1}{n-1}tan\alpha$

$\tan \beta =\frac{n.\sin \alpha \cos \alpha }{1-n{\cos }^2\alpha }$

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

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

$=\frac{\frac{\sin \alpha }{\cos \alpha }+\frac{n.\sin \alpha \cos \alpha }{1-n{\cos }^2\alpha }}{1-(\frac{\sin \alpha }{\cos \alpha }.\frac{n\sin \alpha \cos \alpha }{1-n{\cos }^2\alpha })}$

$=\frac{\frac{\sin \alpha (1-n{\cos }^2\alpha )+n\sin \alpha .{\cos }^2\alpha }{\cos \alpha (1-n{\cos }^2\alpha )}}{\frac{1-n{\cos }^2\alpha -n{\sin }^2\alpha }{(1-n{\cos }^2\alpha )}}$

$=\frac{\sin \alpha -n.\sin \alpha {\cos }^2\alpha +n\sin \alpha .{\cos }^2\alpha }{\cos \alpha .(1-n{\cos }^2\alpha )}\times \frac{(1-n{\cos }^2\alpha )}{1-n({\cos }^2+{\sin }^2\alpha )}$

$=\frac{\tan \alpha }{1-n\left(1\right)}$

$=\frac{\tan \alpha }{1-n}$

$=\frac{\tan \alpha }{-(n-1)}$

$=-\frac{1}{n-1}.\tan \alpha$

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