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zlopas [31]
3 years ago
12

What is the area of the trapezoid shown ?

Mathematics
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

A = 1/2 * (a + c) * h

A = 1/2 * (6 + (2 + 6 + 2)) * 6 = 48

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Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B​
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In order to prove

\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)

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\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)

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\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now, on with the right hand side. We have

1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now simply make this expression one fraction:

1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

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↪Do you have options?

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