If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$, 


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*find $\cos(A+B),\cos(A-B),\sin(A+B)$

*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
 A: $$\begin{align*} \tan A + \tan B &= \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} \\  
             &= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B}  \\
             &= \frac{\sin(A+B)}{\frac{\cos(A+B) + \cos(A-B)}{2}}  \\
             &= \frac{2\sin(A+B)}{\cos(A+B) + \cos(A-B)} \end{align*}$$
A: $$a^2+b^2=2+2\cos(A-B)\implies\cos(A-B)=?$$
$$ab=(\sin A+\sin B)(\cos A+\cos B)=\sin(A+B)+\frac{\sin2A+\sin2B}2$$
$$\implies ab=\sin(A+B)[1+\cos(A-B)]\implies \sin(A+B)=?$$
$$b^2-a^2=\cos2A+\cos2B+2\cos(A+B)=2\cos(A+B)[\cos(A-B)]+1$$
$$\implies\cos(A+B)=?$$
A: From the two given conditions you get $$\tan \left(\frac{A+B}{2}\right)=\frac{a}{b}$$
From that we get $$\cos (A+B)=\frac{b^2-a^2}{b^2+a^2}\\
\sin (A+B)=\frac{2ab}{a^2+b^2}$$
And if you square and add both sides of the two given equations from that you'll get $$\cos (A-B)=\frac{a^2+b^2-2}{2}$$
Now, we get $$\cos^2A+\cos^2B+2\cos A\cos B=b^2\Rightarrow 1+\cos(A+B)\cos(A-B)+2\cos A\cos B=b^2\Rightarrow \cos A\cos B=1/2\left(b^2-1-\frac{(b^2-a^2)(a^2+b^2-2)}{2(a^2+b^2)}\right)$$Hence $$\tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}=\frac{8ab}{2(b^2-1)(a^2+b^2)-(b^2-a^2)(a^2+b^2-2)}=\\\frac{8ab}{(a^2+b^2)^2-4a^2}$$
