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If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)
- a)-25
- b)10
- c)-10
- d)25
Correct answer is option 'A'. Can you explain this answer?
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If tanA and tanB are the roots of the quadratic equation, 3x2 —1...
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If tanA and tanB are the roots of the quadratic equation, 3x2 —1...
The quadratic equation with roots tanA and tanB is given by:
x^2 - (tanA + tanB)x + tanA*tanB = 0
Using the identity tan(A+B) = (tanA + tanB) / (1 - tanA*tanB), we can rewrite the quadratic equation as:
x^2 - 2x(tanA + tanB) + tanA*tanB = 0
Since tanA and tanB are the roots, we can substitute them into the equation:
tanA^2 - 2tanA(tanA + tanB) + tanA*tanB = 0
tanB^2 - 2tanB(tanA + tanB) + tanA*tanB = 0
Simplifying these equations gives:
tanA^2 - 2tanA*tanA - 2tanA*tanB + tanA*tanB = 0
tanB^2 - 2tanB*tanB - 2tanA*tanB + tanA*tanB = 0
tanA^2 - 2tanA^2 - 2tanA*tanB + tanA*tanB = 0
tanB^2 - 2tanB^2 - 2tanA*tanB + tanA*tanB = 0
Simplifying further:
-tanA^2 = 0
-tanB^2 = 0
Therefore, the quadratic equation is:
3x^2 - 2x(tanA + tanB) + tanA*tanB = 0
x^2 - (tanA + tanB)x + tanA*tanB = 0
Using the identity tan(A+B) = (tanA + tanB) / (1 - tanA*tanB), we can rewrite the quadratic equation as:
x^2 - 2x(tanA + tanB) + tanA*tanB = 0
Since tanA and tanB are the roots, we can substitute them into the equation:
tanA^2 - 2tanA(tanA + tanB) + tanA*tanB = 0
tanB^2 - 2tanB(tanA + tanB) + tanA*tanB = 0
Simplifying these equations gives:
tanA^2 - 2tanA*tanA - 2tanA*tanB + tanA*tanB = 0
tanB^2 - 2tanB*tanB - 2tanA*tanB + tanA*tanB = 0
tanA^2 - 2tanA^2 - 2tanA*tanB + tanA*tanB = 0
tanB^2 - 2tanB^2 - 2tanA*tanB + tanA*tanB = 0
Simplifying further:
-tanA^2 = 0
-tanB^2 = 0
Therefore, the quadratic equation is:
3x^2 - 2x(tanA + tanB) + tanA*tanB = 0
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If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. Can you explain this answer?
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If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. 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 If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. Can you explain this answer?.
If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. 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 If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)a)-25b)10c)-10d)25Correct answer is option 'A'. Can you explain this answer?.
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