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.
10 times 'x' is 10 times whatever the value of 'x' is. I hope that makes sense.
11 x 12 =132
i think this is the right answer
Answer:
Given a square ABCD and an equilateral triangle DPC and given a chart with which Jim is using to prove that triangle APD is congruent to triangle BPC.
From the chart, it can be seen that Jim proved that two corresponding sides of both triangles are congruent and that the angle between those two sides for both triangles are also congruent.
Therefore, the justification to complete Jim's proof is "SAS postulate"
Step-by-step explanation: