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Let cos (a + b) = 
and let sin (a – b) = 
, where 0 £ a, b £ 
. Then tan 2a = [AIEEE-2010]
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
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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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? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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