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[(2 sinθ tanθ(1-tanθ)+2 sinθ sec2θ)/(1+tanθ)2]=
- a)[(sinθ)/(1+tanθ)]
- b)[(2sinθ)/(1+tanθ)]
- c)[(2sinθ)/(1+tanθ)2]
- d)[(2sinθ)/(1-tanθ)2]
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
[(2 sinθ tanθ(1-tanθ)+2 sinθ sec2θ)/(1+t...
Explanation:
Given expression:
[(2sinθ tanθ(1-tanθ)+2sinθ sec^2θ)/(1+tanθ)^2]
Expand the expression:
= [(2sinθ tanθ - 2sinθ tan^2θ + 2sinθ sec^2θ)/(1+tanθ)^2]
= [(2sinθ tanθ - 2sinθ tan^2θ + 2sinθ/cos^2θ)/(1+tanθ)^2]
Use trigonometric identities:
- tanθ = sinθ/cosθ
- sec^2θ = 1/cos^2θ
Substitute the identities into the expression:
= [(2sinθ * sinθ/cosθ - 2sinθ * sin^2θ/cosθ + 2sinθ/cos^2θ)/(1+sinθ/cosθ)^2]
= [(2sin^2θ/cosθ - 2sin^3θ/cosθ + 2sinθ/cos^2θ)/(1+sinθ/cosθ)^2]
Simplify the expression:
= [(2sin^2θ - 2sin^3θ + 2sinθ/cosθ)/(1+sinθ/cosθ)^2]
= [(2sinθ(sinθ - sin^2θ + 1/cosθ))/(1+sinθ/cosθ)^2]
= [(2sinθ(1 - sinθ) + 2sinθ/cosθ)/(1+sinθ/cosθ)^2]
= [(2sinθ(1 - sinθ) + 2/cosθ)/(1+sinθ/cosθ)^2]
Further simplify the expression:
= [(2sinθ - 2sin^2θ + 2/cosθ)/(1+sinθ/cosθ)^2]
= [(2sinθ(1 - sinθ) + 2/cosθ)/(1+sinθ/cosθ)^2]
Apply trigonometric identities:
- 1 - sinθ = cosθ
- 1/cosθ = secθ
Substitute the identities into the expression:
= [(2sinθ cosθ + 2secθ)/(1+sinθ/cosθ)^2]
= [(2sinθ cosθ + 2secθ)/(1+ tanθ)^2]
Final expression:
= (2sinθ cosθ + 2secθ)/(1+ tanθ)^2
= (2sinθ/cosθ + 2secθ)/(1+ tanθ)^2
= (2tanθ + 2secθ)/(1+ tanθ)^2
= 2(tanθ + secθ)/(1+ tanθ)^2
= 2sinθ/(1+ tanθ)
Therefore, the correct answer is option B.
Given expression:
[(2sinθ tanθ(1-tanθ)+2sinθ sec^2θ)/(1+tanθ)^2]
Expand the expression:
= [(2sinθ tanθ - 2sinθ tan^2θ + 2sinθ sec^2θ)/(1+tanθ)^2]
= [(2sinθ tanθ - 2sinθ tan^2θ + 2sinθ/cos^2θ)/(1+tanθ)^2]
Use trigonometric identities:
- tanθ = sinθ/cosθ
- sec^2θ = 1/cos^2θ
Substitute the identities into the expression:
= [(2sinθ * sinθ/cosθ - 2sinθ * sin^2θ/cosθ + 2sinθ/cos^2θ)/(1+sinθ/cosθ)^2]
= [(2sin^2θ/cosθ - 2sin^3θ/cosθ + 2sinθ/cos^2θ)/(1+sinθ/cosθ)^2]
Simplify the expression:
= [(2sin^2θ - 2sin^3θ + 2sinθ/cosθ)/(1+sinθ/cosθ)^2]
= [(2sinθ(sinθ - sin^2θ + 1/cosθ))/(1+sinθ/cosθ)^2]
= [(2sinθ(1 - sinθ) + 2sinθ/cosθ)/(1+sinθ/cosθ)^2]
= [(2sinθ(1 - sinθ) + 2/cosθ)/(1+sinθ/cosθ)^2]
Further simplify the expression:
= [(2sinθ - 2sin^2θ + 2/cosθ)/(1+sinθ/cosθ)^2]
= [(2sinθ(1 - sinθ) + 2/cosθ)/(1+sinθ/cosθ)^2]
Apply trigonometric identities:
- 1 - sinθ = cosθ
- 1/cosθ = secθ
Substitute the identities into the expression:
= [(2sinθ cosθ + 2secθ)/(1+sinθ/cosθ)^2]
= [(2sinθ cosθ + 2secθ)/(1+ tanθ)^2]
Final expression:
= (2sinθ cosθ + 2secθ)/(1+ tanθ)^2
= (2sinθ/cosθ + 2secθ)/(1+ tanθ)^2
= (2tanθ + 2secθ)/(1+ tanθ)^2
= 2(tanθ + secθ)/(1+ tanθ)^2
= 2sinθ/(1+ tanθ)
Therefore, the correct answer is option B.
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[(2 sinθ tanθ(1-tanθ)+2 sinθ sec2θ)/(1+tanθ)2]=a)[(sinθ)/(1+tanθ)]b)[(2sinθ)/(1+tanθ)]c)[(2sinθ)/(1+tanθ)2]d)[(2sinθ)/(1-tanθ)2]Correct answer is option 'B'. Can you explain this answer?
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[(2 sinθ tanθ(1-tanθ)+2 sinθ sec2θ)/(1+tanθ)2]=a)[(sinθ)/(1+tanθ)]b)[(2sinθ)/(1+tanθ)]c)[(2sinθ)/(1+tanθ)2]d)[(2sinθ)/(1-tanθ)2]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 [(2 sinθ tanθ(1-tanθ)+2 sinθ sec2θ)/(1+tanθ)2]=a)[(sinθ)/(1+tanθ)]b)[(2sinθ)/(1+tanθ)]c)[(2sinθ)/(1+tanθ)2]d)[(2sinθ)/(1-tanθ)2]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 [(2 sinθ tanθ(1-tanθ)+2 sinθ sec2θ)/(1+tanθ)2]=a)[(sinθ)/(1+tanθ)]b)[(2sinθ)/(1+tanθ)]c)[(2sinθ)/(1+tanθ)2]d)[(2sinθ)/(1-tanθ)2]Correct answer is option 'B'. Can you explain this answer?.
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