Given: 1 sin^2α = 3 sinα cosα

To find: The value of tanα

Proof:

Step 1: Rewrite the given equation using the identity sin^2α = 1 - cos^2α

1 - cos^2α = 3 sinα cosα

Step 2: Rearrange the equation by moving all terms to one side

cos^2α + 3 sinα cosα - 1 = 0

Step 3: Factorize the equation

(cosα + 1)(cosα - 1) + 3 sinα cosα = 0

Step 4: Simplify further

cosα(cosα + 1 - 3 sinα) + (cosα - 1) = 0

Step 5: Divide the equation by cosα

cosα + 1 - 3 sinα + (cosα - 1)/cosα = 0

Step 6: Simplify the equation

1 - 3 sinα + secα - 1 = 0

-3 sinα + secα = 0

Step 7: Rewrite secα in terms of sinα

-3 sinα + 1/cosα = 0

-3 sinα + 1/(√(1 - sin^2α)) = 0

Step 8: Multiply through by (√(1 - sin^2α))

(-3 sinα)(√(1 - sin^2α)) + 1 = 0

-3 sinα√(1 - sin^2α) + 1 = 0

Step 9: Rearrange the equation

1 = 3 sinα√(1 - sin^2α)

1/3 = sinα√(1 - sin^2α)

Step 10: Square both sides of the equation

1/9 = sin^2α(1 - sin^2α)

Step 11: Simplify

9 sin^4α - sin^2α + 1 = 0

Step 12: Notice that the equation is a quadratic in sin^2α

Using the quadratic formula, we can solve for sin^2α:

sin^2α = [1 ± √(1 - 4(9)(1))]/(2(9))

sin^2α = [1 ± √(1 - 36)]/18

sin^2α = [1 ± √(-35)]/18

Since sin^2α cannot be negative, there are no real solutions for sin^2α.

Therefore, the given equation has no real solutions for α.

Hence, it is not possible to determine the value of tanα as the equation is not satisfied.