Question

Plz solve my question class 10 plz proper solve​

Answer 1

EXPLANATION.

⇒ √3sinθ = cosθ.

As we know that,

We can write equation as,

⇒ sinθ/cosθ = 1/√3.

⇒ tanθ = 1/√3.

⇒ tanθ = tan(30°).

⇒ θ = 30°.

To find :

⇒ [(3cos²θ + 2cosθ)/(3cosθ + 2)].

⇒ [cosθ(3cosθ + 2)/(3cosθ + 2)].

⇒ [cosθ].

⇒ cos(30°) = √3/2.

⇒ [(3cos²θ + 2cosθ)/(3cosθ + 2)]. = √3/2.

                                                                                                                   

MORE INFORMATION.

(1) sin²θ + cos²θ = 1.

(2) 1 + tan²θ = sec²θ.

(3) 1 + cot²θ = cosec²θ.

Answer 2

$\huge\sf\dag \: {Given :-}$

$\bigstar \: \boxed{ \sf \: \pink{ \sqrt{3} sin \theta \: = cos \theta}}$

$\sf \leadsto \: \sqrt{3} sin \theta \: = cos \theta \\ \sf \leadsto \: \frac{sin \theta}{cos \theta} \: = \frac{1}{ \sqrt{3} } \: \: \: \: \: \: \\ \sf \leadsto tan \theta \: = \frac{1}{ \sqrt{3} } \: \: \: \: \: \: \: \: \\ \sf \leadsto \: \theta \: = 30 \degree \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\huge\sf \dag{ \: To Find :-}$

$\bigstar \: \boxed{\sf \green{ \frac{3 {cos}^{2} \theta + 2cos \theta }{3cos \theta + 2} }}$

$\huge \sf \dag \: {Solution :-}$

$\sf \dashrightarrow \: \frac{3 {cos}^{2} \theta \: + \: 2cos \theta }{ 3cos \theta \: + 2 } \: \\ \sf\dashrightarrow \: \frac{cos \theta(3 cos \theta \: + 2) }{ 3cos \theta \: + 2} \: \\ \sf \dashrightarrow \: \frac{cos \theta \cancel \gray{(3 cos \theta \: + 2)} }{ \cancel \gray{ 3cos \theta \: + 2} } \: \: \: \\ \sf \dashrightarrow \: cos \theta \: = 30 \degree \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \dashrightarrow \: cos \: 30 \degree \: = \frac{\sqrt{3} }{ 2 } \: \: \: \: \: \:$

$\bigstar \: \boxed{\sf \orange{ \frac{3 {cos}^{2} \theta + 2cos \theta }{3cos \theta + 2 } = 30 \degree}}$