Prove that:
Please solve this problem because tomorrow is my maths exam.

Explanation:

On taking LHS

Cos[(3π/2)+θ]Cos(2π+θ)[Cot{(3π/2)-θ}+Cot(2π+θ)]

We know that

π = 180°

2π = 2×180° = 360°

3π/2 = (3×180°)/2 = 540°/2 = 270°

Now

LHS becomes

Cos(270°+θ)Cos(360°+θ)[Cot(270°-θ)+Cot(360°+θ)]

We know that

Cos (270°+θ) = Sin θ

Cos (360°+θ) = Cos θ

Cot (270°-θ) = Tan θ

Cot (360°+θ) = Cot θ

→ Sin θ Cos θ [Tan θ + Cot θ]

→ Sinθ Cosθ[(Sinθ/Cosθ)+(Cosθ/Sinθ)]

→ Sinθ Cosθ[(Sin²θ+Cos²θ)/(SinθCosθ)]

→ Sin θ Cos θ [1/(Sin θ Cos θ)]

Since Sin²θ+Cos²θ = 1

→ (Sin θ Cos θ)/(Sin θ Cos θ)

→ 1

→ RHS

→ LHS = RHS

Hence, Proved.

Here are the formulae that I have used:

→ π = 180°

→ Cos (270°+θ) = Sin θ

→ Cos (360°+θ) = Cos θ

→ Cot (270°-θ) = Tan θ

→ Cot (360°+θ) = Cot θ

Here are the Trigonometric Identities that I have used:

→ Sin²θ+Cos²θ = 1

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)