Question

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

Answer 1

1 + sin^2 Ø = 3sinØcosØ
divide by cosØ
1/cosØ + tanØsinØ = 3sinØ
secØ + tanØsinØ = 3sinØ
tanØ = (3sinØ - secØ)/sinØ
= 3 - secØ/sinØ
let's stop here:
tanØ = 3 - 1/(sinØcosØ) ******

if tanØ = 1 , sinØ/cosØ = 1 and sinØ = cosØ, Ø = π/2
in *****
LS = tanØ = 1
RS = 3 - 1/sin^2 Ø
= 3 - 1/(1/2) = 3-2 = 1
= LS

if tanØ = 1/2
sinØ = 1/√5 and cosØ = 2/√5

in *****
LS = 1/2
RS = 3 - 1/(sinØcosØ)
= 3 - 1/((1/√5)(2/√5))
= 3 - 1/(2/5)
= 3 - 5/2
= 1/2
= LS

graphic confirmation:
http://www.wolframalpha.com/input/?i=solve+tanx+%3D+3+-+1%2F%28sinxcosx%29

Answer 2

or, you can try it like this

1 + sin^2 Ø = 3sinØcosØ
sin^2 Ø = (1-cos2Ø)/2

1 + (1-cos2Ø)/2 = 3/2 sin2Ø
3 - cos2Ø = 3sin2Ø
9 - 6cos2Ø + cos^2 2Ø = 9 - 9cos^2 2Ø
10cos^2 2Ø + 6cos2Ø = 0
2cos2Ø(5cos2Ø-3) = 0

cos2Ø = 0 or cos2Ø = 3/5

so Ø = π/4 as above and tanØ=1, or

tanØ = √((1-cos2Ø)/(1cos2Ø))
= √(1 - 3/5)/(1 + 3/5) = √(1/4) = 1/2

Those are the angles in the 1st quadrant that satisfy the equation.