If `1+sin^2 theta= 3 sintheta cos theta,` then prove that `tan theta = 1` or `tan theta =1/2`

Explanation

`1+sin^(2)theta=3sin thetacos theta rArr sec^(2)theta+tan^(2)theta=3 tan theta " "["dividing throughout by "cos^(2)theta]`
`rArr (1+tan^(2)theta)+tan^(2)theta=3tantheta`
`rArr 2tan^(2)theta-3tan theta+1=0`
`rArr 2tan^(2)theta-2tantheta-tantheta+1=0`
`rArr 2tantheta(tantheta-1)-(tan theta-1)=0`
`rArr (tantheta-1)(2tantheta-1)=0`
`rArr tan theta =1 or (1)/(2).`