If sin (A + B) = cos (A + B), what is the value of tan A?

$\frac{1-tanB}{1+tanB}$

According to question ,

sin (A + B) = cos (A + B)

by using formula

sin A cos B + cos A sin B = cos A cos B - sin A sin B

on rearranging ,

sin A cos B + sin A sin B = cos A cos B - cos A sin B

sin A(cos B + sin B) = cos A (cos B - sin B)

on rearranging , 

\(\frac{sinA}{cosA}\)  =  \(\frac{cos B - sin B}{cos B + sin B}\) 

taking out common cosB from left hand side

tanA = \(\frac{1 - sin B / cosB}{1  + sin B/ cosB}\)


tan A = \(\frac{1 - tanB}{1+ tan B}\)