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Harlamova29_29 [7]
3 years ago
12

Use trigonometric identities to simplify each expression. 1/cot^2 (x) - 1/cos^2(x)

Mathematics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

- 1

Step-by-step explanation:

\frac{1}{\cot ^2\left(x\right)}-\frac{1}{\cos ^2\left(x\right)}\\\\\mathrm{Use\:the\:basic\:trigonometric\:identity}:\quad \frac{1}{\cos \left(x\right)}=\sec \left(x\right),\\\frac{1}{\cot ^2\left(x\right)}-\sec ^2\left(x\right)\\\\\mathrm{Use\:the\:basic\:trigonometric\:identity}:\quad \frac{1}{\cot \left(x\right)}=\tan \left(x\right),\\\tan ^2\left(x\right)-\sec ^2\left(x\right)\\\\=> \frac{\sin ^2\left(x\right)}{\cos ^2\left(x\right)}-\frac{1}{\cos ^2\left(x\right)}\\

=> \frac{\sin ^2\left(x\right)-1}{\cos ^2\left(x\right)}\\\\=> -\frac{\cos ^2\left(x\right)}{\cos ^2\left(x\right)}\\=> - 1

Hope  that  helps!

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zloy xaker [14]

Answer:

Number of​ restaurant-purchased meals eaten in a​ restaurant = 70

Number of​ restaurant-purchased meals eaten in a​ car = 22

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Step-by-step explanation:

Let the number of people that will eat in a restaurant be R.

Let the number of people that will eat in a car be C.

Let the number of people that will eat in a home be H.

From the information given in the problem we have,

R+C+H = 152 ____equation (1)

C+H-R = 12 ____equation (2)

R = H+10 ____equation (3)

1) Plugging in R=H+10 from equation 3 into the equation 2, we get

C+H-R=12

=> C+H-(H+10)=12

=> C+H-H-10=12

2) Cancelling out +H and -H, we get

C-10=12

3) Add 10 to both sides

C-10+10=12+10

4) Cancelling out -10 and +10, we get

C=22

5) Plugin C=22 in equation 1, we have

R+C+H = 152

=> R+22+H=152

Subtract 22 from both sides,

R+22+H-22=152-22

Cancelling out +22 and -22 from the left side, we get

R+H=130 ____let it be equation (4)

6) Plugin C=22 in equation 2, we have

C+H-R = 12

22+H-R = 12

Subtracting 22 from both the sides, we get

22+H-R-22 = 12-22

Cancelling out +22 and -22 from the left side,

H-R = -10 ____let it be equation (5)

7) Adding equation 4 and equation 5, we get

(R+H)+(H-R) = 130 + (-10)

=> R+H+H-R = 130-10

8) Cancelling out R and -R from the left side, we get

2H = 120

9) Dividing both sides by 2, we get

\frac{2H}{2} = \frac{120}{2}

10) Cancelling out the 2's from the left side, we have

H=60

11) Plugging in C=22 and H=60 in the equation 1, we have

R+C+H = 152

=> R+22+60=152

=> R + 82 = 152

12) Subtracting 82 from both the sides, we get

R + 82 -82 = 152 -82

13) Cancelling out +82 and -82 from the left side, we get

R = 70

<em>So, C=22, H=60, R=70</em>

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Answer:

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