Question

Answer please no wrong answer wrong answer will be reported​

Answer 1

Step-by-step explanation:

Solution :-

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.

Used Formulae:-

→ π = 180°

→ Cos (270°+θ) = Sin θ

→ Cos (360°+θ) = Cos θ

→ Cot (270°-θ) = Tan θ

→ Cot (360°+θ) = Cot θ

Used Trigonometric Identities:-

→ Sin²θ+Cos²θ = 1