Find the value of $$\cos (A + B)$$, if $$\tan A = \frac{3}{4}$$ and $$\tan B = \frac{5}{12}$$ and $$A$$ and $$B$$ are in quadrant I.

Using this formula ⇒tan(a+b)=$$\frac{tanA+tanB}{1-tanA.tanB}$$

⇒then substitute tana and tanb values 

⇒$$\frac{3/4+5/12}{1-3/4×5/12}$$

 tan(a+b)=$$\frac{56}{33}$$

To find the cos (a+b)

Use this formula $$\sec^2a$$=1+$$\tan^2a$$

Where in formula replace a by a+b

$$\sec^2(a+b)$$=1+$$\tan^2(a+b)$$

                       =1+($$\frac{56}{33})^2$$

                       =$$\frac{4225}{1089}$$    

    $$\sec^2(a+b)$$=  ($$\frac {65}{33})^2$$

Therefore sec(a+b)=$$\frac{65}{33}$$

And we know that The reciprocal of cosine function is secant

Hence cos(a+b)=$$\frac{33}{65}$$