My school mysterious trigonometry test question

So I was working on my math test and the whole class and I couldn't do 2 numbers from the test. I ask another class they couldn't do it either. Until I found the answer from my study group, I ask how did they got those answer, but all of them said, someone gave me the answer. All study groups that I joined have the same answer. So I'm really confused, and am wondering if those question really have the same answer?

These are the questions:

Number 1

$\tan(A)+\tan(B)=\sqrt 6$

$\cos(A)\cos(B)=\frac{1}{6}\sqrt 3$

The question is find the value of $\tan(A+B)$

Number 2

$\sin(X+Y)=A\sin(X-Y)$

Find the value of $(\frac{A+1}{A-1})\tan(Y)$

Here are my attempts:

Number 1

$\tan(A+B)=\frac{\sqrt6}{1-\left(\frac{\sin(A)\sin(B)}{(\frac{1}{6})\sqrt 3}\right)}$

Number 2 I can't do it at all

Solution 1:

Hint for Q1

$$\\ \tan(A)+\tan(B) \\= \frac{\sin(A)}{\cos(A)}+\frac{\sin(B)}{\cos(B)} \\= \frac{\sin(A)\cos(B)+\cos(B)\sin(A)}{\cos(A)\cos(B)} \\= \frac{\sin(A+B)}{\frac{1}{6}\sqrt{3}}$$

And with $\sin(A+B)$, you can directly work out $\tan(A+B)$ by finding $\cos(A+B)$ first.

Hint for Q2

$$A = \frac{\sin(X+Y)}{\sin(X-Y)}\\ A+1 = \frac{\sin(X+Y)+\sin(X-Y)}{\sin(X-Y)} = \frac{2\sin(X)\cos(Y)}{\sin(X-Y)}\\ A-1 = \frac{\sin(X+Y)-\sin(X-Y)}{\sin(X-Y)} = \frac{2\cos(X)\sin(Y)}{\sin(X-Y)}\\$$

With these, I think you can then work it out, since I simplified the expression down.

Solution 2:

1)As highlighted @Prometheus, you have $$\sqrt{6} = \tan(A)+\tan(B)=\frac{\sin(A+B)}{\sqrt{3}/6}$$ so instead of figure out how to get $\cos(A+B)$, for better solve the trigonometric equation for angle $A+B$. To do this, from above you got that $$\sin(A+B)=\frac{\sqrt{18}}{6}=\frac{\sqrt{2}}{2}$$ so $A+B=\frac{\pi}{4}$ or $\frac{3\pi}{4}$. Then $\tan(A+B)=1$.

2)The second one is easier for you after the Hint provided by Prometheus.