Question

Cos²(45°+theta)+cos²(45°-theta)/tan(60°+theta)tan(30°-theta) + (sin47°/cos43°)-2cos²45°

Answer 1

Answer:

1

Step-by-step explanation:

N =  Cos²(45°+theta)+cos²(45°-theta)/tan(60°+theta)tan(30°-theta) + (sin47°/cos43°)-2cos²45°

let theta = ∝

N =  Cos²(45°+∝)+cos²(45°-∝)/tan(60°+∝)tan(30°-∝) +(sin47°/cos43°)-2cos²45°

N =  Cos²(45°+∝)+cos²(45°-∝)/tan(60°+∝ )tan(30°-∝) +(sin47°/sin47°)-2cos²45°

N =  Cos²(45°+∝)+cos²(45°-∝)/tan(60°+∝ )tan(30°-∝) + 1 - 1

N =  Cos²(45°+∝)+cos²(45°-∝)/tan(60°+∝ )cot(60°+∝)

N =  Cos²(45°+∝)+cos²(45°-∝)/1

N =  Cos²(45°+∝)+cos²(45°-∝)

N =  Cos²(45°+∝)+sin²(45°+∝)

N = 1