The answer is the last one (absolute value)
Answer:
Verified below
Step-by-step explanation:
We want to show that (Cos2θ)/(1 + sin2θ) = (cot θ - 1)/(cot θ + 1)
In trigonometric identities;
Cot θ = cos θ/sin θ
Thus;
(cot θ - 1)/(cot θ + 1) gives;
((cos θ/sin θ) - 1)/((cos θ/sin θ) + 1)
Simplifying numerator and denominator gives;
((cos θ - sin θ)/sin θ)/((cos θ + sin θ)/sin θ)
This reduces to;
>> (cos θ - sin θ)/(cos θ + sin θ)
Multiply top and bottom by ((cos θ + sin θ) to get;
>> (cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ)
In trigonometric identities, we know that;
cos 2θ = (cos² θ - sin²θ)
cos²θ + sin²θ = 1
sin 2θ = 2sinθcosθ
Thus;
(cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ) gives us:
>> cos 2θ/(1 + sin 2θ)
This is equal to the left hand side.
Thus, it is verified.
X-60 WITH X BEING THE BANK ACCOUNT TOTAL - THE WITHDRAWAL DO U HAVE ANY ANSWER CHOICES?
Nitrogen Radius = 5.8 x 10⁻¹¹ m
Beryllium Radius = 1.12 x 10⁻¹⁰ m
Let's find the quotient of N/Be :
(5.8x10⁻¹¹)/(1.12x10⁻¹⁰). But 10⁻¹¹/10⁻¹⁰ = 10⁽⁻¹¹⁺¹⁰⁾ = 10⁻¹ = 1/10 = 0.1
→ (5.8/1.12).(0.1) = 0.58/1.12 = 0.518.
Conclusion: the radius of Be is almost double than the radius of N
