<u>Explanation</u><u>:</u>
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
<u>Hence, Proved.</u>
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
