# Find A+B when tanA+tanB=1 and cosA*cosB=(3^(1/2))/2.?

Oct 18, 2017

Given $\tan A + \tan B = 1$ and $\cos A \cdot \cos B = \frac{\sqrt{3}}{2}$,

we are to find out A+B=?

Now
$\tan A + \tan B = 1$

$\implies \sin \frac{A}{\cos} A + \sin \frac{B}{\cos} B = 1$

$\implies \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = 1$

$\implies \sin A \cos B + \cos A \sin B = \cos A \cos B$

$\implies \sin \left(A + B\right) = \frac{\sqrt{3}}{2} = \sin \left(\frac{\pi}{3}\right)$

$\implies A + B = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{3} \text{ where } n \in \mathbb{Z}$

Oct 18, 2017

In the interval $\left[0 , 2 \pi\right)$, we have $A + B = \frac{\pi}{3}$ or $\frac{2 \pi}{3}$

and general solution is $A + B = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{3}$

#### Explanation:

$\tan A + \tan B = \sin \frac{A}{\cos} A + \sin \frac{B}{\cos} B = 1$

or $\frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = 1$

or $\sin \frac{A + B}{{3}^{\frac{1}{2}} / 2} = 1$

or $\sin \left(A + B\right) = \frac{\sqrt{3}}{2} = \sin \left(\frac{\pi}{3}\right)$

and $A + B = \frac{\pi}{3}$ or $\frac{2 \pi}{3}$ in the interval $\left[0 , 2 \pi\right)$

and general solution is $A + B = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{3}$