Finding $\cot(\beta)$, knowing $\sin(\alpha+\beta)=4/5$ and $\sin(\alpha-\beta)=3/5$ Let's assume that for $0<\beta<\alpha<\frac{\pi}{2}$, $\sin(\alpha+\beta) = \frac{4}{5}$, and $\sin(\alpha-\beta) = \frac{3}{5}$. Then, how could we find $\cot(\beta)$?
$$\sin(\alpha+\beta)+\sin(\alpha-\beta) = 2\sin(\alpha)\cos(\beta) = \frac{7}{5}$$
$$\sin(\alpha+\beta)-\sin(\alpha-\beta) = 2\sin(\beta)\cos(\alpha) = \frac{1}{5}$$
$$\tan(\alpha)\cot(\beta) = 7$$
But I am not sure where this would lead us.
 A: $\displaystyle sin( a+b) =4/5\ \ \ sin( a-b) =3/5.$
$\displaystyle So\ cos( a+b) \ =\ 3/5$ but that means $\displaystyle sin\left(\frac{\pi }{2} -( a+b)\right) =sin( a-b) =3/5$
So $ $$\displaystyle \frac{\pi }{2} =a+b+a-b$
Or $\displaystyle a\ =\ \frac{\pi }{4}$
So $cot(a) = 1 $
And since you have already figured out $tan(a)cot(b)=7$ so $cot(b)=7$
A: $$
\begin{aligned}
& \sin ^2(\alpha+\beta)+\sin ^2(\alpha-\beta)=\frac{16}{25}+\frac{9}{25}=1 \\
\Rightarrow \quad & \sin ^2(\alpha+\beta)=1-\sin ^2(\alpha-\beta)=\cos ^2(\alpha-\beta)\\ \Rightarrow \quad & 
(\sin \alpha \cos \beta+\sin \beta \cos \alpha)^2=(\cos \alpha \cos \beta+\sin \alpha \sin \beta)^2 \\\Rightarrow \quad &\left(\cos ^2 \beta-\sin ^{2} \beta\right)\left(\cos ^2 \alpha-\sin ^{2} \alpha\right)=0 \\ \Rightarrow \quad 
&\cos ^2 \beta=\sin ^2 \beta \quad \textrm{  or }\cos ^2 \alpha=\sin ^2 \alpha \\\Rightarrow \quad & \cot\beta =1 \textrm{ or } \tan \alpha =1 \\ \Rightarrow \quad & \cot \beta=1 \quad \textrm{ or } \quad  7 \quad \textrm{ (By }\tan \alpha \cot \beta = 7)
\end{aligned}
$$
