1.

If `cos(alpha+beta)=(4)/(5) and sin(alpha-beta)=(5)/(13)` , where `alpha` lie between 0 and ` (pi)/(4)`, then find that value of `tan2alpha`.

Answer» Given that, ` cos(alpha+beta)=(4)/(5) and sin(alpha-beta)=(5)/(13)`
`rArr" "sin(alpha+beta)=sqrt(1-(16)/(25))=sqrt((9)/(25))= pm (3)/(5)`
`therefore" "sin(alpha+beta)=(3)/(5) `
and `" "cos(alpha-beta)=sqrt(1-(25)/(169))=sqrt((144)/(169))=pm(12)/(13) `
`" " cos(alpha-beta)=(12)/(13)`
Now, `tan(alpha+beta)=(sin(alpha+beta))/(cos(alpha+beta))" "`[since, `alpha` lies between 0 and ` (pi)/(4)`
`((3)/(5))/((4)/(5))=(3)/(4)`
and `" "tan(alpha-beta)=(sin(alpha-beta))/(cos(alpha-beta))=((5)/(13))/((12)/(13))=(5)/(12)`
`therefore" "tan2alpha= tan(alpha+beta+alpha-beta)`
` " "=(tan(alpha+beta)+ tan (alpha-beta))/(1- tan(alpha+ beta)*tan(alpha-beta))" "[becausetan(xpmy)=(tanxpmtany)/(1pmtanx*tany)] `
`" "=((3)/(4)+(5)/(12))/(1-(3)/(4)*(5)/(12))=((9+5)/(12))/((16-5)/(16))=(14xx16)/(12xx11)=(56)/(33)`


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