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.
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Answered by
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sinA = 15/17 so cosA gonna be 8/17 and tanA = 15/8
cosB= 12/13 therefore sinB= 5/13 and tanB = 5/12
now sin(A+B)= sinAcosB + sinBcosA = 15/17*12/13 + 8/17*5/13 = 220/221
similarly cos(A-B)= cosAcosB + sinAsinB = 8/17*12/13 + 15/17*5/13 = 171/221
and now
tan(A+B)= tanA + tanB / 1- tanAtanB = 15/8 + 5/12 /1- 15/8*5/12 =220/21
hope my answer is right
even if its wrong please manage, its must be calculation mistake
anyways through the formula nad get ur answer
cosB= 12/13 therefore sinB= 5/13 and tanB = 5/12
now sin(A+B)= sinAcosB + sinBcosA = 15/17*12/13 + 8/17*5/13 = 220/221
similarly cos(A-B)= cosAcosB + sinAsinB = 8/17*12/13 + 15/17*5/13 = 171/221
and now
tan(A+B)= tanA + tanB / 1- tanAtanB = 15/8 + 5/12 /1- 15/8*5/12 =220/21
hope my answer is right
even if its wrong please manage, its must be calculation mistake
anyways through the formula nad get ur answer
Answered by
1
Step-by-step explanation:
Correct Answer-
220/221
171/221
220/21
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