Question

If sinA=15/17 and cosB=12/13, find the values of sin(A+B), cos(A-B) and tan(A+B), A,B are the positive acute angles.

Answer 1

sinA=15/17  and cosB=12/13
Using, sin²A+cos²A=1 we get,
cos²A=1-sin²A=1-225/289=(289-225)/289=64/289
cosA=8/17 (neglecting the neggative sign since A,B are positive acute angles)
sin²B=1-cos²B=1-144/169=(169-144)/169=25/169
sinB=5/13 (neglecting the neggative sign since A,B are positive acute angles)
sin(A+B)
=sinAcosB+cosAsinB
=15/17×12/13+8/17×5/13
=180/221+40/221
 =220/221
 cos(A-B)
 =cosAcosB-sinAsinB
=8/17×12/13-15/17×5/13
=96/221-75/221
 =21/221
 tanA=sinA/cosA=(15/17)/(8/17)=15/8 and
tanB=sinB/cosB=(5/13)/(12/13)=5/12   
tan(A+B)
=(tanA+tanB)/(1-tanAtanB)
=(15/8+5/12)/(1-15/8×5/12)
=[(45+10)/24]/[(96-75)/96]
=(55/24)/(21/96)
= 55/24×96/21
=210/21
=10