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3 years ago

Using trig identities, prove the equation shown below.

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Answer:

Step-by-step explanation:

Prove that

sec(x) - sin(x)tan(x)=cosx

sec(x)=1/cos(x) tan(x)=sin(x)/cos(x)

1/cos(x) - sin(x) x sin(x)/cos(x)

(1-sin(x) x sin(x))/cos(x)

(1-sin^2(x))/cos(x) 1-sin^2(x)=cos^2(x)

Cos^2(x)/cos(x)

(Cos(x) x cos(x))/cos(x)

cos(x) proved

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I need this asap ty:p

Anastasy [175]

Answer:

(a) x = -2y

Step-by-step explanation:

You can tell if an equation is a direct variation equation if it can be written in the format y = kx.

Note that there is no addition and subtraction in this equation.

Let's put these equations in the form y = kx.

(a) x = -2y

y = x/-2 → y = -1/2x
This is equivalent to multiplying x by -1/2, so this is an example of direct variation.

(b) x + 2y = 12

2y = 12 - x
y = 6 - 1/2x
This is not in the form y = kx since we are adding 6 to -1/2x. Therefore, this is NOT an example of direct variation.

-2y = -3x
y = 3/2x
This follows the format of y = kx, so it is an example of direct variation.

(d) 5x² + y = 0

y = -5x²
This is not in the form of y = kx, so it is NOT an example of direct variation.

(e) y = 0.3x + 1.6

1.6 is being added to 0.3x, so it is NOT an example of direct variation.

(f) y - 2 = x

y = x + 2
2 is being added to x, so it is NOT an example of direct variation.

The following equations are examples of direct variation:

x = -2y
3x - 2y = 0

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