grenade
Last Activity: 9 Years ago
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1) 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) = 56/33
plz aprve