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If 12sin2θ + 4cos2θ= 6 , 0° < θ < 90° than the value of tan2θ is? |
(\(\frac{2}{\sqrt {3}}\))2 (\(\frac{1}{\sqrt {3}}\))2 0 1 |
(\(\frac{1}{\sqrt {3}}\))2 |
θ = 30° satisfying the equation. So, put θ = 30° tan2θ = tan2 30° = (\(\frac{1}{\sqrt {3}}\))2 |
