Question

If `sinA=3/5, cosB=-12/13`, where `A in 0,pi/2[,B in]pi,(3pi)/2`,then evaluate the following:
i) `sin(A-B)`
ii) `cos(A-B)`
iii) `tan(A+B)`

Answer 1

`sinA=3/5`, A lies in 1st quadrant.
`therefore cos^(2)A=1-sin^(2)A=1-(3/5)^(2)=16/25`
`rArr cosA=4/5`
and `tanA=(sinA)/(cosA)=(3/5)/(4/5)=3/4`
Again `cosB=-12/13`, B lies in 3rd quadrant.
`therefore sin^(2)B=-5/13` (In 3rd quadrant, sin is negative)
`therefore tanB=(sinB)/(cosB)=(-5/13)/(-12/13)=5/12`
i) `sin(A-B)=sinAcosB-cosAsinB`
`=3/5(-12/13)-4/5(-5/13)`
`=(-36+20)/(65)=-16/65`. Ans.
ii) `cos(A-B)=cosA.cosB+sinAsinB`
`=4/5(-12/13)+3/5(-5/13)`
`=(-48-15)/(65)=-63/65` Ans.
iii) `tan(A+B)=(tanA+tanB)/(1-tanAtanB)`
`=(3/4+5/12)/((1-3/4.5)/12)=56/33`. Ans.