Question

if 1+sin20=3sin0cos0 , then prove that tan0=1 or tan0=1/2
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Answer 1

$\underline{\textsf{Proof :}}$

$\mathsf{Now,\:1+sin^{2}\theta=3\:sin\theta\:cos\theta}$

$\mathsf{Dividing\:both\:sides\:by\:cos^{2}\theta\:,\,cos\theta \neq 0}$

$\to \mathsf{\frac{1+sin^{2}\theta}{cos;{2}\theta}=\frac{3\:sin\theta\:cos\theta}{cos^{2}\theta}}$

$\to \mathsf{\frac{1}{cos^{2}\theta}+\frac{sin^{2}\theta}{cos^{2}\theta}=3\:\frac{sin\theta}{cos\theta}}$

$\to \mathsf{sec^{2}\theta+tan^{2}\theta=3\:tan\theta}$

$\to \mathsf{1+tan^{2}\theta+tan^{2}\theta=3\:tan\theta}$

$\to \mathsf{2\:tan^{2}\theta-3\:tan\theta+1=0}$

$\to \mathsf{2\:tan^{2}\theta-2\:tan\theta-tan\theta+1=0}$

$\to \mathsf{2\:tan\theta(tan\theta-1)-1(tan\theta-1)=0}$

$\to \mathsf{(tan\theta-1)(2\:tan\theta-1)=0}$

$\mathsf{Either\:tan\theta-1=0\:or,\:2\:tan\theta-1=0}$

$\implies \boxed{\mathsf{tan\theta=1\:,\:tan\theta=\frac{1}{2}}}$

$\textsf{Hence, proved.}$