Question

If 1+sin2θ = 3sinθ cosθ, then show 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

Given  1+ sin 2θ = 3 sinθ cosθ
            1 + 2 sinθ cosθ = 3 sinθ cosθ
            =>  sinθ cosθ = 1         , divide by cos θ  both sides
            =>  tan θ = sec² θ
            =>  tanθ  = 1 + tan² θ
tan² θ - tanθ + 1 = 0
tanθ =  [ 1 + - √(1-4) ] /2 

As discriminant is < 0,  it is not possible.  There is no θ for which the given question is valid.
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So the question probably means  1 + sin² θ = 3 sinθ cosθ

=> we divide both sides by cos² θ..
=>  sec²θ + tan²θ = 3 tan 
=> 2 tan²θ + 1 - 3 tanθ = 0
=>  tan θ=  [3 +- √(9-8) ] /4
              = 1 or 1/2