SINA=15/17 AND COSB=12/13
THAN THE VALUE OF tan(A+B)IS
OPTIONS ARE GIVE AS..
A.220/221
B.171/221
B.220/21
D.171/21
Answers
Answered by
1
sinA=15/17
so, cosA=√(1-sin^2A)
cosA=√{(289-225)/289
cosA=√64/225=8/17
and, cosB=12/13
so, sinB=√{(1)-(12/13)^2}
sinB={(169-144)/169}
sinB=5/13
tanA=sinA/cosA=(15/17)÷(8/17)=15/8
tanB=sinB/cosB=(5/13)÷(12/13) = 5/12
tan(A+B)=(tanA+tanB)÷(1-tanA.tanB)
tan(A+B)=(15/8+5/12)÷(1-15*5/8*12)
= (45+10)/24÷ (96-75)/96
tan(A+B)=220/21
so, cosA=√(1-sin^2A)
cosA=√{(289-225)/289
cosA=√64/225=8/17
and, cosB=12/13
so, sinB=√{(1)-(12/13)^2}
sinB={(169-144)/169}
sinB=5/13
tanA=sinA/cosA=(15/17)÷(8/17)=15/8
tanB=sinB/cosB=(5/13)÷(12/13) = 5/12
tan(A+B)=(tanA+tanB)÷(1-tanA.tanB)
tan(A+B)=(15/8+5/12)÷(1-15*5/8*12)
= (45+10)/24÷ (96-75)/96
tan(A+B)=220/21
shubh1:
THANKS BRO
Similar questions