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1. If $1 + \sin^{2} θ = 3 \sin θ . \cos θ$ , then prove that $\tan θ = 1 o r \frac{1}{2} .$

Dear Student,
Let, theta=A

$1 + \sin^{2} A = 3 \sin A \cos A \Rightarrow (\sin^{2} A + \cos^{2} A) + \sin^{2} A = 3 \sin A \cos A \Rightarrow \cos^{2} A + 2 \sin^{2} A = 3 \sin A \cos A D i v i d e b y \cos^{2} A w e g e t 1 + 2 \tan^{2} A = 3 \tan A L e t \tan A = t {2 t}^{2} - 3 t + 1 = 0 \Rightarrow (2 t - 1) (t - 1) = 0 \Rightarrow t = \frac{1}{2} o r 1 \tan A = 1 o r \frac{1}{2}$

Regards.

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