Question

$\rm\:Prove\:that\:tanx=1\:or\:\dfrac{1}{2}\:if\:1+sin^2x=3 sinx cosx}$

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: 1 + {sin}^{2}x \: = \: 3 \: sinx \: cosx$

Divide both sides by $\rm \: cos^2x$, we get

$\rm \: \dfrac{1}{ {cos}^{2} x} + \dfrac{ {sin}^{2}x }{ {cos}^{2}x } = \dfrac{3sinxcosx}{ {cos}^{2} x}$

$\rm \: {sec}^{2}x \: + \: {tan}^{2}x = 3tanx$

$\rm \: 1 + {tan}^{2}x \: + \: {tan}^{2}x = 3tanx$

$\rm \: 2{tan}^{2}x + 1= 3tanx$

$\rm \: 2{tan}^{2}x - 3tanx + 1= 0$

$\rm \: 2{tan}^{2}x - 2tanx - tanx + 1= 0$

$\rm \: 2tanx(tanx - 1) - 1(tanx - 1) = 0$

$\rm \: (tanx - 1)(2tanx - 1) = 0$

$\rm \: tanx - 1= 0 \: \: \: or \: \: \: 2tanx - 1 = 0$

$\bf\implies \:tanx = 1 \: \: or \: \: tanx = \dfrac{1}{2}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\rm \: tanx = \dfrac{sinx}{cosx}$

$\rm \: secx = \dfrac{1}{cosx}$

$\rm \: {sec}^{2}x = 1 + {tan}^{2}x$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \\ \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$