Practice your tangent sum and difference formula!

We will use the following elementary identities:

• $\tan(x\pm y)=\dfrac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}$
• $(x+y)(x-y)=x^2-y^2$

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Take $x=\tan(2A)$ and $y=\tan(B)=\dfrac{1}{3}$ . Using the above-mentioned identities, we get $x=\dfrac{4}{3}$ and also get that,

$\tan(2A+B)\tan(2A-B)=\frac{x^2-y^2}{1-x^2y^2}=\frac{27}{13}$

Note that only the option $3\textrm{ and }\frac{9}{13}$ satisfies this result. Hence, that is our required answer out of the options. Observing this actually becomes trivial once you note that $13$ is a prime and $\gcd(27,13)=1$ .

P.S. : I hope this covers the "Bonus" in your problem.

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Of course, one can manually calculate both of the values mentioned, but then again, that's too much work and we are lazy, so..... :3

Given„ tanA=1/2 & tanB=1/3.

Tan2A=tan(A+A)= (2tanA)/[1-(tanA)^2] =4/3.

Hence, tan(2A+B)=3.

Similarly, we can calculate tan(2A-B) by using formula.. Which is 9/13.