cos a= 4/5 and cos B= 12/13 what's tan(a+b) and tan (a-b)

Answer 1

You will need to use your formulas for tan(a+b) and tan(a-b) which will require you to figure out the values of tan a and tan b

If cos a =4/5 (adjacent/hypotenuse) then by pythagorean’ theorem opposite is 3 and tan a =3/4

If cos b= 12/13 the tan a =5/13

tan (a+b) = tana +tan b/ 1-tanatanb =

(3/5+ 5/13)/ 1-( 3/5 x 5/13)

(64/65) / 1-15/65

(64/65)/ 50/65= 64/50=32/25

tan(a-b) =tana -tanb/ 1+tanatanb

I will leave the calculation to you

Answer 2

Without knowing what quadrant A and B are in (quadrant 1 or quadrant 4; knowing cos is positive in each case restricts us to these quadrants), we cannot answer this question.

We know that |sin A| = 3/5 and |sin B| = 5/13, implying that |tan A| = 3/4 and |tan B| = 5/12

tan(A + B) = (tan A + tan B)/(1 - tan A tan B); tan(A - B) = (tan A - tan B)/(1 + tan A tan B)