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cos(a+b)=4/5,sin(a-b)=5/13 and a,b lie between 0 and pi/4, then tan 2a=
Cos(A+B)=4/5= 0.8 A+B =36.87 Sin(A-B)=5/13 = 0.3846 A-B = 22.62 by adding 2A =36.87 + 22.62 = 59.49 tan 2A = tan 59.49= 1.6969plz aprve
or1) Since both A & B are in (0, π/4), both (A + B) & (A - B) will be only in (0, π/2); hence tan(A+B) & tan(A-B) both will be positive. 2) As cos(A+B) = 4/5, tan (A+B) = 3/4. [Applying right triangle rule, adj = 4; hyp = 5; ==> opp = 3] Similalry as sin(A-B) = 5/13, tan(A-B) = 12/13 3) Tan(2A) = tan(A + A + B - B) = tan{(A + B) + (A - B)} = [{tan(A+B) + tan(A-B)}/{1 - tan(A+B)*tan(A-B)}] [Since, identity, tan(x+y) = {tan(x) + tan(y)}/{1 - tan(X)*tan(y)}] Substituting the numerical values, Tan(2A) = (3/4 + 5/12)/{1 - (3/4)*(5/12)} = 56/33 Thus tan(2A) = 56/33plz aprve
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