Question

if root 3 sin theta - cos theta = 2 then find the value of theta​

Answer 1

Answer:

The answer is here

Step-by-step explanation:

It is given that \sqrt{3}sin\theta=cos{\theta}

3

sinθ=cosθ

⇒tan{\theta}=\frac{1}{\sqrt{3}}tanθ=

3

1

⇒{\theta}=30^{\circ}θ=30

∘

Thus, the value of the given expression is:

\frac{3cos^2{\theta}+2cos{\theta}}{3cos{\theta}+2}

3cosθ+2

3cos

2

θ+2cosθ

=\frac{3cos^230^{\circ}+2cos30^{\circ}}{3cos30^{\circ}+2}

3cos30

∘

+2

3cos

2

30

∘

+2cos30

∘

=\frac{3(\frac{3}{4})+2(\frac{\sqrt{3}}{2})}{3(\frac{\sqrt{3}}{2})+2}

3(

2

3

)+2

3(

4

3

)+2(

2

3

)

=\frac{9+4\sqrt{3}}{2(3\sqrt{3}+4)}

2(3

3

+4)

9+4

3

which is the required solution.

Answer 2

Step-by-step explanation:

$\sqrt{3} \sin( \theta) - \cos( \theta) = 2 \\ = > \frac{ \sqrt{3} }{2} \sin( \theta) - \frac{1}{2} \cos( \theta) = \frac{2}{2} \\ = > \sin( \theta) \cos(30) - \cos( \theta) \sin(30) = 1 \\ = > \sin( \theta - 30) = \sin(90) \\ = > \theta - 30 = 90 \\ = > \theta = 120$