Question

If 1 + sin2θ = 3sinθ cosθ , then prove that tanθ = 1 or 1/2

Answer 1

Solution –
 
1 + sin^2θ = 3 sinθ cosθ
 
Divide both side of the equation with cos^2θ
 
1/cos^2θ + sin^2θ/cos^2θ = 3 sinθ/cosθ
 
sec^2θ + tan^2θ = 3tanθ
 
1 + tan^2θ + tan^2θ = 3tanθ
 
1 + 2tan^2θ – 3tanθ = 0
 
Substitute tanθ = a
 
2a^2 – 3a + 1 = 0
 
Solve the quadratic equation to find out the roots.
 
2a^2 – 2a – a + 1 = 0
 
2a (a – 1) – 1 (a – 1) = 0
 
(2a – 1) ( a – 1) = 0
 
2a – 1 = 0 and (a – 1) = 0
 
2a = 1 and a = 1 a = 1/2 and a = 1
 
Hence a = tanθ
 
tanθ = 1/2 and tanθ = 1

Answer 2

1 + Sin² Ф = 3 sinФ  cosФ
Multiply both sides by sec²Ф 

sec² Ф + sin² Ф /cos² Ф = 3 sinФ / cosФ
1 + tan² Ф + tan²Ф = 3 tan Ф
2 tan² Ф - 3 tan Ф + 1 = 0

we can use factorization by factors :  2 X 1  =>  - 2 - 1  = -3 
  so 2 tan² Ф - 2 tanФ - tan Ф + 1 = 0
      (2 tan Ф -1)  (tan Ф - 1)  = 0
       tan Ф = 1/2  or  1

or  by quadratic equation method:
tan Ф = [ 3 + - √(9 - 8)] / 4
          = 1 or 1/2