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Mathematics

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SIZIF [17.4K]3 years ago

8 0

Answer:

though I told him he is in love you to be able and he will have a test done to you guys at the hospital and the other

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Prove that {(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)

USPshnik [31]

First, expand the terms inside the bracket you will get

$(( \tan {}^{2} (x) + 2 \tan(x) \sin(x) + \sin {}^{2} (x) - ( \tan {}^{2} (x) - 2 \tan(x) + \sin {}^{2} (x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$( 4 \tan(x) \sin(x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16 \tan {}^{2} (x) \sin {}^{2} (x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16 \tan {}^{2} (x) (1 - \cos {}^{2} (x) ) = 16 (\tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16( \tan {}^{2} (x) - \frac{ \sin {}^{2} (x) \cos {}^{2} ( {x}^{} ) }{ \cos {}^{2} (x) }$

$16( \tan {}^{2} (x) - \sin {}^{2} (x) ) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

5 0

2 years ago

Find f(-2) for the function: f(x)= 3x^2-1, x<1 x+2, x> 1

azamat

F(x) = 3x² - 1
f(-2) = 3(-2)² - 1
f(-2) = 3(4) - 1
f(-2) = 12 - 1
f(-2) = 11

f(x) = x < 1
f(-2) = -2 < 1

f(x) = x + 2
f(-2) = -2 + 2
f(-2) = 0

f(x) = x > 1
f(-2) = -2 > 1
f(-2) = -2 < 1

8 0

3 years ago

Find the area of the figure. area units2

Verdich [7]

Answer:

$33 units^{2}$