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r-ruslan [8.4K]
2 years ago
10

Prove that sec^2θcosec*2θ-2-cot^2θ=tan2θ

Mathematics
1 answer:
Tema [17]2 years ago
5 0

Answer:

Step-by-step explanation:

I think you mean tan^2θ, not tan2θ.

Use the trigonometric identities

secθ = 1/cosθ

cscθ = 1/sinθ

tanθ = sinθ/cosθ

sin²θ = 1-cos²θ

cos²θ = 1-sin²θ

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HELPPP
aivan3 [116]

Answer:

681ft^2

Step-by-step explanation:

Start by finding the area of the whole thing with the missing triangle:

33 x 22 = 726

Then find the area of the triangle. You can do this by first finding out the side lengths of the triangle:

33 - 24 = 9 and 22 - 12 = 10

Then multiply both sides and divide by two because it is a triangle:

10 x 9 = 90       90/2 = 45

Then subtract the area of the triangle from the rectangle:

726 - 45 = 681

5 0
2 years ago
Read 2 more answers
An=-31+(n-1)-11 find the eleventh term
Lilit [14]

An=-31+(n-1)-11 find the eleventh term

a11 = -31 + (-11) (11-1)

a11 = -31 +-11(10)

a11 =-31 +-110

a11 = -142

3 0
3 years ago
alexandrea and imani each think of a number. alexandras number is 8 more than Imanis number. the sum of two numbers is 50. what
juin [17]
Alexandera=a
imani=i

a is 8 more than i or
a=8+i

the sum of the 2 numbers is 50 or
a+i=50

a=8+i
so subsitute 8+i for a in a+i=50
8+i+i=50
8+2i=50
subtract 8 from both sides
2i=42
divide both sides by 2
i=21

subsiutte into a=8+i

a=8+21
a=29



a=29
i=21
3 0
3 years ago
Which set of rectangular coordinates describes the same location as the polar coordinates (6sqrt 2, 3pi/4)
Sonja [21]
Polar coordinates=<span>(6sqrt 2, 3pi/4)=(r, theta)→r=6 sqrt 2, theta=3pi/4
Rectangular coordinates=(x,y)=?

x=r cos theta=(6 sqrt 2) cos(3pi/4)=(6 sqrt 2)(-sqrt 2 / 2)
x=-3 (sqrt 2)^2=-3(2)→x=-6

y=r sin theta=(6 sqrt 2) sin (3pi/4)=(6 sqrt 2)(sqrt 2 / 2)
y=3 (sqrt 2)^2=3(2)→y=6

Rectangular coordinates of the point = (x,y) =(-6,6)

Answer: Option a. (-6,6) </span>
8 0
3 years ago
A rectangle has a height of 4 and a width of x^2+3x+2
Mashutka [201]
Base times height 4x^2+12x+8
8 0
3 years ago
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