Question

$\mathbb\purple{QUESTION:-}$


​

​





$\begin{gathered}\begin{gathered}{\large \qquad \boxed{\boxed{\begin{array}{cc} \hookrightarrow \sf \: If \: 2 - cos ^{2} \theta = 3 \: sin \theta \: cos \theta, \: \\ \\ \sf \: sin \theta \neq \: cos \theta .\: So, \: What \\ \\ \sf will \: tan \theta? \end{array}}}}\end{gathered}\end{gathered}$




$\bigstar\bf{ \: Remember \: this :-}$
$\rm\longmapsto{No \: spam}$


$\rm\longmapsto{ Only \: good \: answer \: with \: explanation}$

​

Answer 1

Step-by-step explanation:

• Given,2-cos²θ=3sin θ cos θ Dividing both sides by cos²θ, we get

2 sec² θ-1=3 tan θ

⇒2(1+tan²θ)-1=tanθ

⇒2+2tan²θ-1=3tanθ

⇒2tan²θ-3tan θ(tan θ+1)=0

⇒2tan²θ-2tan θ-tan θ+1=0

⇒2tan θ(tan θ-1)-(tan θ-1)=0

⇒(2tan θ-1) (tan θ-1)=0

Hope that helps u❤

Answer 2

EXPLANATION.

⇒ 2 - cos²θ = 3sinθcosθ.

⇒ sinθ ≠ cosθ.

As we know that,

Divide both equation by cos²θ, we get.

⇒ [2 - cos²θ/cos²θ] = [3sinθcosθ/cos²θ].

⇒ [2/cos²θ - cos²θ/cos²θ] = [3sinθ/cosθ].

⇒ [2sec²θ - 1] = [3 tanθ].

⇒ 2sec²θ - 1 = 3tanθ.

As we know that,

Formula of :

⇒ 1 + tan²θ = sec²θ.

Using this formula in the equation, we get.

⇒ 2(1 + tan²θ) - 1 = 3tanθ.

⇒ 2 + 2tan²θ - 1 = 3tanθ.

⇒ 1 + 2tan²θ = 3tanθ.

⇒ 2tan²θ - 3tanθ + 1 = 0.

Factorizes the equation into middle term splits, we get.

⇒ 2tan²θ - 2tanθ - tanθ + 1 = 0.

⇒ 2tanθ(tanθ - 1) - 1(tanθ - 1) = 0.

⇒ (2tanθ - 1)(tanθ - 1) = 0.

⇒ tanθ = 1/2   and   tanθ = 1.