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