Question

if 2-cos^2θ=3sinθcosθ,where sinθ ≠ cosθ the value of tanθ is​

Answer 1

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

Given that,

$\rm \: sin\theta \ne \: cos\theta \: \rm\implies \:tan\theta \: \ne \: 1$

and

$\rm \: 2 - {cos}^{2}\theta = 3sin\theta cos\theta$

can be rewritten as

$\rm \: 1 + 1- {cos}^{2}\theta = 3sin\theta cos\theta$

We know,

$\boxed{ \rm{ \: {sin}^{2}\theta + {cos}^{2}\theta = 1 \: }}$

So, using this, we get

$\rm \: 1 + {sin}^{2}\theta = 3sin\theta cos\theta$

can be further rewritten as

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

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

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

We know,

$\boxed{ \rm{ \: {sec}^{2}\theta - {tan}^{2}\theta = 1 \: }}$

So, using this, we get

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

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

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

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

$\rm \: 2tan\theta(tan\theta - 1) - 1(tan\theta - 1) = 0$

$\rm \: (tan\theta - 1)(2tan\theta - 1) = 0$

$\rm\implies \:2tan\theta - 1 = 0 \: \: as \: tan\theta \: \ne \: 1$

$\rm\implies \:\boxed{ \bf{ \:tan\theta \: = \: \frac{1}{2} \: \: }}$

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

Additional Information :-

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

Answer 2

$\huge\blue{ANSWER:-}$

Given, 2−cos²θ=3sinθcosθ

Dividing both sides by cos²θ, we get

2sec²θ−1=3tanθ

2(1+tan²θ)−1=3tanθ

⇒2+2tan²θ−1=3tanθ

⇒2tan²θ−3tanθ+1=0

⇒2tan²θ−2tanθ−tanθ+1=0

⇒2tanθ(tanθ−1)−(tanθ−1)=0

⇒(2tanθ−1)(tanθ−1)=0

⇒tanθ=1/2 and 1

I hope you have understood the ans yaar❤️