Find the value of cos(A +B) if tan A = 3/4 and tan B = 5/12 and A and B are in quadrant I.

​$$\tan A = \frac{3}{4} \\ \ \\\tan B = \frac{5}{12}$$​​

Both A and B are in Quadrant I, meaning all trigonometric ratios are positive.

​$$\cos(A + B) = \cos A \cos B - \sin A \sin B \\ \ \\\tan \theta = \frac{\sin \theta}{\cos \theta} \\ \ \\\sin^2 \theta + \cos^2 \theta = 1$$​​

​$$\tan A = \frac{\sin A}{\cos A} = \frac{3}{4}$$​​

So,

​$$\sin A = 3k \quad \text{and} \quad \cos A = 4k$$​​

Using $$\sin^2 A + \cos^2 A = 1$$​​

​$$(3k)^2 + (4k)^2 = 1 \\ \ \\ k^2 + 16k^2 = 19 \\ \ \\ 25k^2 = 1 \quad \Rightarrow \quad k = \frac{1}{5}$$​​

So:

​$$\sin A = \frac{3}{5} \quad \text{and} \quad \cos A = \frac{4}{5}$$​​

Now,

​$$\tan B = \frac{\sin B}{\cos B} = \frac{5}{12} ​\sin B = 5m \quad \text{and} \quad \cos B = 12m$$​​

Using , $$\sin^2 B + \cos^2 B = 1$$​​$$(5m)^2 + (12m)^2 = 1 \\ \ \\ 25m^2 + 144m^2 = 1 \\ \ \\169m^2 = 1 \quad \Rightarrow \quad m = \frac{1}{13}$$​

So:

​$$\sin B = \frac{5}{13} \quad \text{and} \quad \cos B = \frac{12}{13}$$​​

Now,

​$$\cos(A + B) = \cos A \cos B - \sin A \sin B \\ \ \\ \cos(A + B) = \left(\frac{4}{5} \times \frac{12}{13}\right) - \left(\frac{3}{5} \times \frac{5}{13}\right) \\ \ \\ \cos(A + B) = \frac{48}{65} - \frac{15}{65} \\ \ \\ \cos(A + B) = \frac{33}{65}$$​​

Thus, the value of $$\cos(A + B)cos(A+B) \quad\text{ is} \quad \frac{33}{65}​.$$​​