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

Given, $$\tan \beta = \frac{n\sin\alpha\cos\alpha}{1-n\cos^2\alpha}$$ Then $\tan(\alpha + \beta)$ is equal to

1. $(n-1)\tan\alpha$
2. $(n+1)\tan\alpha$
3. $\frac{\tan\alpha}{n+1}$
4. $\frac{\tan\alpha}{1-n}$

I would also appreciate practical methods of tackling problems such as this in competitive examinations.

• Do you know of the addition formula for the tangent function? – mickep Jan 28 '16 at 20:06
• @mickep You mean: $$\tan(A+B)=\frac{\tan A + \tan B}{1-\tan A\tan B}$$? – Hungry Blue Dev Jan 28 '16 at 20:08

I think it will be easier if one first simplifies $\tan\beta$, $$\tan\beta=\frac{n\sin\alpha\cos\alpha}{1-n\cos^2\alpha}=\frac{n\tan\alpha}{1/\cos^2\alpha-n}=\frac{n\tan\alpha}{\tan^2\alpha+1-n}.$$ Next, use the addition formula for $\tan(\alpha+\beta)$ and insert the expression above everywhere you encounter $\tan\beta$. Simplify, and you got your result.
A quicker method would be to substitute a simple value for $\alpha$ such as $\frac{\pi}{4}$ and do the same compound angle formula simplification