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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)

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Something is wrong. I calculated

\begin{align} \sin A &= \dfrac{\sqrt 3}{2}\sin B \\ \cos A &= \dfrac{\sqrt 5}{3}\cos B \\ \sin^2 A + \cos^2 A &= \dfrac 34 \sin^2 B + \dfrac59 \cos^2 B\\ 1 &= \dfrac 34 - \dfrac{7}{36} \cos^2 B\\ \cos^2 B = -\dfrac{9}{7} \end{align}

Which is impossible.

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HINT:

$$\tan A=\frac{\sin A}{\cos A}=\frac{\frac{\sqrt{3}}2\sin B}{\frac{\sqrt{5}}3\cos B}=\frac{3\sqrt{3}}{2\sqrt{5}}\tan B\tag{1}$$ And $$1=\sin^2 A+\cos^2 A=\frac{3}{4}\sin^2 B+\frac{5}{9}\cos^2 B=\frac59\left(\sin^2 B+\cos^2B\right)+\frac7{36}\sin^2 B$$ Last equation implies $$\sin^2B=\frac{16}7$$

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$$\frac{|\sin A|}{|\sin B|}=\frac{\sqrt{3}}{2} <1 \iff \frac{|\cos A|}{|\cos B|}>1$$

But $$\frac{|\cos A|}{|\cos B|}=\frac{\sqrt{5}}{3}<1$$

that's impossible.

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Hint:

Let $\dfrac{\sin A}{\sqrt3}=\dfrac{\sin B}2=p$ and $\dfrac{\cos A}{\sqrt5}=\dfrac{\cos B}3=q$

$\implies(\sqrt3p)^2+(\sqrt5q)^2=1$

and $(2p)^2+(3q)^2=1$

Solve for $p^2,q^2$

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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.

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