Prove that: \cos( \frac{3\pi}{2} + \theta) \cos(2\pi + \theta) [ \cot( \frac{3\pi}{2} - \theta) + \cot(2\pi + \theta) ] = 1
Please solve this problem because tomorrow is my...

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

Hope this helps!!

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles