We have proven that the trigonometric identity [(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] equals 1 + (secθ * cosec θ)
<h3>How to solve Trigonometric Identities?</h3>
We want to prove the trigonometric identity;
[(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] = 1 + sec θ
The left hand side can be expressed as;
[(tan θ)/(1 - (1/tan θ)] + [(1/tan θ)/(1 - tan θ)]
⇒ [tan²θ/(tanθ - 1)] - [1/(tan θ(tanθ - 1)]
Taking the LCM and multiplying gives;
(tan³θ - 1)/(tanθ(tanθ - 1))
This can also be expressed as;
(tan³θ - 1³)/(tanθ(tanθ - 1))
By expansion of algebra this gives;
[(tanθ - 1)(tan²θ + tanθ.1 + 1²)]/[tanθ(tanθ(tanθ - 1))]
Solving Further gives;
(sec²θ + tanθ)/tanθ
⇒ sec²θ * cotθ + 1
⇒ (1/cos²θ * cos θ/sin θ) + 1
⇒ (1/cos θ * 1/sin θ) + 1
⇒ 1 + (secθ * cosec θ)
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Answer:
14.478
Step-by-step explanation:
search in anything 5.7 in centimeters boom your answer
Solving a system of equations, we will see that there are 6000 kg of coal on the pile.
<h3>
How many kilograms of coal are in the pile of coal?</h3>
Let's say that there are N kilograms of coal, and it is planned to burn it totally in D days, these are our variables.
We know that:
- If the factory burns 1500 kilograms per day, it will finish burning the coal one day ahead of planned.
- If it burns 1000 kilograms per day, it will take an additional day than planned.
Then we can write the system of equations:
(1500)*(D - 1) = N
(1000)*(D + 1) = N
Because N is already isolated in both sides, we can write this as:
(1500)*(D - 1) = N = (1000)*(D + 1)
Then we can solve for D:
(1500)*(D - 1) = (1000)*(D + 1)
1500*D - 1500 = 1000*D + 1000
500*D = 2500
D = 2500/500 = 5
Now that we know the value of D, we can find N by replacing it in one of the two equations:
(1500)*(D - 1) = N
(1500)*(5 - 1) = N
(1500)*4= N = 6000
This means that there are 6000 kilograms of coal on the pile.
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Answer:
204 hours
Step-by-step explanation:
3264 divided by 16 = 204 hours
By Evaluating the Compound Interest, we come to know that Rajesh will have enough money in the account to cover all of the required loan payments.
The Principal Amount(P) = $30,000
Rate of Interest (r) = 2.16 %
Time(t) = 10 years
Number of Times it is Compounded in a year(n) = 12
Now, we have
Putting all the values, we evaluate the amount,
Hence, the Amount after Compound Interest = $37,225.87
Now, The loan amount that he pays = 300 *12*10 = $ 36,000
Yes, he will have enough money in the account to cover all of the required loan payments.
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