If $\tan x + \tan y = 4$ and $\cos x + \cos y = 1/5$, find $\tan(x+y)$.
Well, from the first condition, we get
$$\tan x + \tan y = \frac{\sin(x+y)}{\cos x \cos y}=4 \implies \sin(x+y)=4\cos x \cos y$$
Then,
$$\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{4}{1-\frac{\sin x \cos y}{\cos x \cos y}}=\frac{4}{1-\frac{4\sin x \cos y}{\sin(x+y)}}$$
But in this way, I couldn't get use of the second condition. Actually, I squared the first condition and replaced $\tan^2$ with $\sec^2-1$. Then, squared the second condition to find something useful about $\sec^2$'s. Yet, nothing that works.
Therefore, started substituting $\cos x = a$ and $\cos y = b$ to get
$$
\begin{cases}
\cfrac{\pm\sqrt{1-a^2}}{a} + \cfrac{\pm\sqrt{1-b^2}}{b} = 4 \\
a + b = \cfrac{1}{5}
\end{cases}
$$
Nonetheless, this system doesn't seem to be as easy as this problem might be given as a multiple-choice problem (2-3 minutes for solving).
Any help is appreciated.
Best Answer
I show my attempts: I wrote down the system \begin{align} &\frac{s_1}{c_1}+\frac{s_2}{c_2}=4,\\ &c_1+c_2=1/5,\\ &c_1^2+s_1^2=1,\\ &c_2^2+s_2^2=1 \end{align} solved with Wolfram Mathematica and obtained $8$ solution, but by symmetry they are only $4$. Excluded a complex solution, I have three solutions.
Substituted in $$ \frac{\frac{s_1}{c_1}+\frac{s_2}{c_2}}{1-\frac{s_1s_2}{c_1c_2}} $$ I get the three values $$ (-1.47878, 0.0385584, 0.209877). $$
Furthermore, if it's a multiple solution test, with proposed results $r_1,\ldots,r_n,$ then you can proceed backward: find the product of tangents as $$ p=1-\frac{4}{r_i} $$ then given sum $s$ of tangents and their product, find the values of tangents $t_1,t_2$ solving $$ t^2-s t+p=0 $$ and finally calculate the sum of cosines as $$ \cos(\arctan t_1)+\cos(\arctan t_2) $$ that should give $1/5$ for a correct solution.