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If ( 1 sin²A ) = 3sinAcosA then prove that tanA = 1 or 1/2?
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If ( 1 sin²A ) = 3sinAcosA then prove that tanA = 1 or 1/2?
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If ( 1 sin²A ) = 3sinAcosA then prove that tanA = 1 or 1/2?
 

Given:

(1 + sin²A) = 3sinAcosA



To prove:

tanA = 1 or 1/2



Proof:



Simplifying the given equation:

(1 + sin²A) = 3sinAcosA

1 + sin²A = 3sinAcosA

sin²A + 1 = 3sinAcosA

sin²A + cos²A = 3sinAcosA

1 = 3sinAcosA



Using trigonometric identity:

sin²A + cos²A = 1



Comparing the above equation with the simplified equation:

sin²A + cos²A = 3sinAcosA



Equating the coefficients:

1 = 3sinAcosA



Dividing both sides by cosA:

secA = 3sinA



Dividing both sides by sinA:

tanA = 3



Therefore, tanA = 3.



Now, we need to find the possible values of A for which tanA = 3.



Solving for A:

tanA = 3

A = tan⁻¹(3)



Using a calculator, we find:

A ≈ 71.57 degrees



Therefore, one possible value of A is approximately 71.57 degrees.



Now, we need to find the other possible value of A.



Using the periodicity of the tangent function:

tan(A + 180 degrees) = tanA



Therefore, the other possible value of A is:

A + 180 degrees ≈ 71.57 + 180 degrees

A + 180 degrees ≈ 251.57 degrees



Therefore, the possible values of A are approximately 71.57 degrees and 251.57 degrees.



Now, we need to find tanA for these values of A:



tan(71.57 degrees) ≈ 3

tan(251.57 degrees) ≈ 3



Therefore, the possible values of A for which tanA = 3 are approximately 71.57 degrees and 251.57 degrees.



However, we need to find the values of A for which tanA = 1 or 1/2.



Now, we need to find the possible values of A for which tanA = 1 or 1/2.



Solving for A when tanA = 1:

A = tan⁻¹(1
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