If 1+sin2α=3sinαcosα, then values of cotα are . . .

Question : If 1+sin²α=3sinαcosα, then values of cotα are

a) -1, 1

b) 0, 1

c) 1, 2

d) -1, -1

Doubt by Nimit & Pushkar

Solution :

1+sin²α=3sinαcosα
sin²α+cos²α+sin²α=3sinαcosα
[∵ sin²θ+cos²θ=1]
2sin²α+cos²α=3sinαcosα
2sin²α+cos²α-3sinαcosα=0
2sin²α-3sinαcosα+cos²α=0
2sin²α-[2sinαcosα+sinαcosα]+cos²α=0

2sin²α-2sinαcosα-sinαcosα+cos²α=0
2sinα[sinα-cosα]-cosα[sinα-cosα]=0
[sinα-cosα][2sinα-cosα]=0
sinα-cosα=0
sinα=cosα
sinα/cosα = 1
tanα = 1
1/cotα = 1
cotα=1

2sinα-cosα = 0
2sinα=cosα
2sinα/cosα = 1
tanα = 1/2
1/cotα = 1/2
cotα = 2

∴ cotα = 1 or 2

Hence, c) 1,2 , would be the correct option.