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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.

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  • $\begingroup$ Do you know of the addition formula for the tangent function? $\endgroup$ – mickep Jan 28 '16 at 20:06
  • $\begingroup$ @mickep You mean: $$\tan(A+B)=\frac{\tan A + \tan B}{1-\tan A\tan B}$$? $\endgroup$ – Hungry Blue Dev Jan 28 '16 at 20:08
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Since you knew about the addition formula for tan, here is a push in the right direction:

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.

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  • $\begingroup$ Thank you. Your solution worked for me. But I was wondering if there was any quicker way to do it... (assumptions and short-cuts allowed). $\endgroup$ – Hungry Blue Dev Jan 28 '16 at 20:36
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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

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  • $\begingroup$ I agree that is a bit simpler, given that one of the four alternatives really is correct. $\endgroup$ – mickep Jan 29 '16 at 6:26

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