Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $\tan B$ but that doesn't help in getting the value of $\tan A + \tan B$. The value comes in terms of $\tan A$ or $\tan B$ but the expected answer is independent of any variable .
Also
$$\frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \tan(A) + \tan(B)$$
We could get a value only if instead of $\cos A$ there was $\sin B$ in the relation(which we get on adding the ratios)

Although there is something wrong with this question,there is a way which I think maybe a little bit easier to solve this kind of problem.
   $$\tan A + \tan B = \frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \frac{{\sin A \over \sin B}+ {\cos A \over \cos B}}{\cos A \over \sin B} $$
then to get the value of $\cos A \over \sin B $ :
   $$ {\sin A \over \sin B }= {\sqrt{3} \over 2}  \Rightarrow {(1-\cos^2 A) \over \sin^2 B} = {3 \over 4} \Rightarrow {1 \over \cos^2A} = 1+{{3\cdot \sin^2 B} \over 4\cdot \cos^2A }$$
   $${\cos A \over \cos B }= {\sqrt{5} \over 3}  \Rightarrow {\cos^2A \over (1-\sin^2 B)} = {5 \over 9} \Rightarrow {1 \over \cos^2A} = {9 \over 5}+{{\sin^2 B} \over \cos^2A }$$
then
$$ {\sin^2B \over  \cos^2A  }={-16 \over 5} $$
this is impossible.
But if the question is right and then you can get the value of $\cos A \over \sin B $ ,and just get the  answer with the first equation.
