If cos(A+B)=4/5 , sin(A-B)=5/13 , and A,B lies between 0 and 45 , then tan2A=?

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Solution

given,

cos(A+B)=4/5, thus tan(A+B)=3/4 from triangle

sin(A-B)=5/13,thus tan(A-B)=5/12.

then tan(2A)=tan((A+B)+(A-B))

=(tan(A+B)+tan(A-B))/(1-tan(A+B)tan(A-B))

=(3/4+5/12)/(1-(3/4)(5/12))

= 56/33.

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