Step-by-step explanation:
We must prove that
cos²a(csc²a-cot²a) = cos²a
If we look at both sides, we can see that we have cos²a * something = cos²a. Therefore, if we can get that something to equal 1, we have our proof. In this case, that something is csc²a-cot²a. Using this information, we can work from within the parenthesis and go from there.
We can start by expanding the items in the parenthesis. Taking that csc(x) = 1/sin(x) and cot(a) = cos(x)/sin(x), we can say that
cos²a(csc²a-cot²a) = cos²a(1/sin²a - cos²a/sin²a). Because both items in the parenthesis have a denominator of sin²a, we can subtract cos²a from 1 to get
cos²a(1/sin²a - cos²a/sin²a)= cos²a((1-cos²a)/sin²a))
Next, we know that cos²a+sin²a=1, so 1-cos²a = sin²a. Plugging that in, we get
cos²a((1-cos²a)/sin²a)) = cos²a(sin²a/sin²a)
= cos²a(1)
= cos²a
Answer:
$2370.59
Step-by-step explanation:
the final balance is $2370.59
Answer:
m∠QRS = 52°
Step-by-step explanation:
From the given figure,
In right triangles ΔRQT and ΔRST,
QT = TS = 7.5 units [Given]
RT ≅ RT [Reflexive property]
ΔRQT ≅ ΔRST [By HL theorem of congruence]
Therefore, m(∠QRT) = m(∠SRT) = 26° [CPCTC]
m(∠QRS) = 2m(∠QRT)
= 2×(26°)
= 52°
Answer:
$28360
Step-by-step explanation:
Step one:
Given data
P=20000
rate= 6%
time= 6years
Step two:
compound interest
A=P(1+r)^t
substitute
A=20000(1+0.06)^6
A=20000(1.06)^6
A=20000*1.418
A=$28360
Answer:
The probability of of a randomly chosen student being exactly 21 years old.
= 1.293
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given Population size n = 500</em>
<em>Mean of the Population = 20 years and 6 months</em>
<em> = </em><em></em>
<em>Standard deviation of the Population = 2 years</em>
Let 'X' be the range of ages of the students on campus follows a normal distribution
Let x =21
<em>The probability of a randomly chosen student being exactly 21 years old.</em>
<em>P( Z≤21) = 0.5 + A( 0.2) </em>
= 0.5 +0.793
= 1.293