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 Let cos (a + b) =  and let sin (a – b) = , where 0 £ a, b £ . Then tan 2a =      [AIEEE-2010]
  • a)
      
  • b)
     
  • c)
  • d)
     
Correct answer is option 'B'. Can you explain this answer?
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Let cos (a + b) =and let sin (a –b) =, where 0 £ a, b &pou...
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 
hopes it help you
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Let cos (a + b) =and let sin (a –b) =, where 0 £ a, b £. Then tan 2a = [AIEEE-2010]a)b)c)d)Correct answer is option 'B'. Can you explain this answer?
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