JEE Exam > JEE Questions > 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^...
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2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)?
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2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2...
1st step: 2/3 【 tan^-1 [ (3ab^2/b^3) -( a^3/b^3) ] /[ (b^3/b^3 )- (3a^2b/b^3) ] 】 + 2/3 【 tan^-1 [(3xy^2/y^3) -( x^3/y^3)]/ [( y^3/y^3) -( 3xy^2/y^3)]】 = tan^-1 【 (2alpha beta / alpha^2) / [(alpha^2/alpha^2) - (beta^2/alpha^2)]】2nd step: 2/3 tan^-1[{ 3(a/b) - (a/b)^3 } / (1-3(a/b)^2)]+ 2/3 tan^-1[ { 3(x/y) - (x/y)^3 } /( 1-3 (x/y)^2)] = tan^-1【 { 2beta/alpha} / 1- {beta/ alpha}^2】3rd step:2/3 { 3tan^-1 (a/b) } + 2/3 {3tan^-1 (x/y)} = 2tan^-1 {beta/alpha}4th step:tan^-1(a/b) + tan^-1 (x/y) = tan^-1 (beta/ alpha)
Community Answer
2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2...
Given equation:
2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b) = 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2αβ/α^2 - β^2)
Proof:
To prove the given equation, we will evaluate both sides separately and show that they are equal.
Step 1: Evaluate the left-hand side (LHS)
LHS = 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b)
Step 2: Simplify the expression inside the tangent function
Let's simplify the expression inside the tangent function:
3ab^2 - a^3/b^3 - 3a^2b
Step 3: Factor out common terms
We can factor out 'a' from the expression:
a(3b^2 - a^2/b^3 - 3ab)
Step 4: Simplify the expression further
By rearranging the terms, we get:
a(3b^2 - 3ab - a^2/b^3)
Step 5: Rewrite the expression using the difference of squares formula
We can rewrite the expression as:
a[(√3b)^2 - 2(√3b)(a/√b^3) + (a/√b^3)^2 - (a/√b^3)^2]
Step 6: Apply the tangent addition formula
Using the tangent addition formula, we have:
tan(A + B) = (tanA + tanB) / (1 - tanA*tanB)
In this case, A = tan^-1(√3b) and B = tan^-1(a/√b^3)
Step 7: Apply the tangent addition formula to the expression
Using the tangent addition formula, we can rewrite the expression as:
a[(tan(tan^-1(√3b)) + tan(tan^-1(a/√b^3))) / (1 - tan(tan^-1(√3b))*tan(tan^-1(a/√b^3)))]
Step 8: Simplify the expression using the properties of tangent
The tangent of an inverse tangent is simply the argument itself:
a[(√3b + (a/√b^3)) / (1 - (√3b)*(a/√b^3))]
Step 9: Simplify further by multiplying numerator and denominator by √b^3
a[(√3b√b^3 + a) / (√b^3 - (√3b)a)]
Step 10: Simplify the expression one last time
By simplifying the expression, we get:
a(√3b√b^3 + a) / (√b^3 - (√3b)a)
Step 11: Evaluate the right-hand side (RHS)
RHS = tan
2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b) = 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2αβ/α^2 - β^2)
Proof:
To prove the given equation, we will evaluate both sides separately and show that they are equal.
Step 1: Evaluate the left-hand side (LHS)
LHS = 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b)
Step 2: Simplify the expression inside the tangent function
Let's simplify the expression inside the tangent function:
3ab^2 - a^3/b^3 - 3a^2b
Step 3: Factor out common terms
We can factor out 'a' from the expression:
a(3b^2 - a^2/b^3 - 3ab)
Step 4: Simplify the expression further
By rearranging the terms, we get:
a(3b^2 - 3ab - a^2/b^3)
Step 5: Rewrite the expression using the difference of squares formula
We can rewrite the expression as:
a[(√3b)^2 - 2(√3b)(a/√b^3) + (a/√b^3)^2 - (a/√b^3)^2]
Step 6: Apply the tangent addition formula
Using the tangent addition formula, we have:
tan(A + B) = (tanA + tanB) / (1 - tanA*tanB)
In this case, A = tan^-1(√3b) and B = tan^-1(a/√b^3)
Step 7: Apply the tangent addition formula to the expression
Using the tangent addition formula, we can rewrite the expression as:
a[(tan(tan^-1(√3b)) + tan(tan^-1(a/√b^3))) / (1 - tan(tan^-1(√3b))*tan(tan^-1(a/√b^3)))]
Step 8: Simplify the expression using the properties of tangent
The tangent of an inverse tangent is simply the argument itself:
a[(√3b + (a/√b^3)) / (1 - (√3b)*(a/√b^3))]
Step 9: Simplify further by multiplying numerator and denominator by √b^3
a[(√3b√b^3 + a) / (√b^3 - (√3b)a)]
Step 10: Simplify the expression one last time
By simplifying the expression, we get:
a(√3b√b^3 + a) / (√b^3 - (√3b)a)
Step 11: Evaluate the right-hand side (RHS)
RHS = tan
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2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)?
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2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)?.
2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)?.
Solutions for 2/3tan^-1(3ab^2 - a^3/b^3 - 3a^2b). + 2/3tan^-1(3xy^2 - x^3/y^3 - 3x^2y) = tan^-1(2alpha beta/ alpha^2 - beta^2)? in English & in Hindi are available as part of our courses for JEE.
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