Question

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

Answer 1

$1 + {sin}^{2} A = 3sinAcosA \\ \\</p><p>dividing \: throughout \: by \: {cos}^{2}A \\ \\</p><p>\frac{1}{{cos}^{2}A} + \frac{{sin}^{2}A}{{cos}^{2}A} = \frac{3sinAcosA}{{cos}^{2} A} \\</p><p>{sec}^{2} A + {tan}^{2} A = 3tanA$

... $(\frac{1}{cosA} = secA ; \frac{sinA}{cosA} = tanA)$

1 + tan²A + tan²A = 3tanA

... (sec²A = 1 + tan²A)

2tan²A - 3tanA + 1 = 0

2tan²A - 2tanA - tanA + 1 = 0

2tanA(tanA - 1) - 1(tanA - 1) = 0

(2tanA - 1)(tanA - 1) = 0

2tanA - 1 = 0 or tanA - 1 = 0

tanA = ½ or tanA = 1

... Hence Proved!