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wariber [46]
3 years ago
8

Verify sinx/1-cosx + sinx/1+cosx = 2cscx

Mathematics
1 answer:
Andru [333]3 years ago
3 0
   
\displaystyle  \\ 
 \frac{\sin x}{1-\cos x} + \frac{\sin x}{1+\cos x} = \\  \\ 
 =\frac{\sin x(1+\cos x)}{(1-\cos x)(1+\cos x)} + \frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)} = \\  \\ 
 =\frac{\sin x+\sin x\cos x}{1-\cos^2x} + \frac{\sin x-\sin x\cos x}{1-\cos^2x} = \\  \\ 
 =\frac{\sin x+\sin x\cos x + \sin x-\sin x\cos x}{1-\cos^2x} = \\  \\ 
=\frac{2\sin x }{\sin^2x} = \frac{2 }{\sin x} =2 \times \frac{1 }{\sin x} = 2 \times \csc x =  \boxed{2\csc x }



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Answer: (A) 30%

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Given : The probability that a bird will lay an egg =0.75

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Which of the following is an equation for a line that is NOT parallel to the other three?
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The equation of the line that is not parallel to others is (C) 2x-3y= -1

<h3>How to determine the equation of the line is not parallel?</h3>

From the question, we have the following parameters that can be used in our computation:

A) 3x-2y=5

B) -3x+2y=9

C) 2x-3y= -1

D) -3x+2y=8

Make y the subject in equations

So, we have the following representation

A) y = 3/2x - 5/2

B) y = 3/2x + 9/2

C) y= 2/3x + 1/3

D) y = 3/2x + 4

As a general rule of linear equations

Parallel lines have equal slope

A linear equation is represented as

y = mx + c

Where

slope = m

Using the above as a guide, we have the following:

A) y = 3/2x - 5/2 ⇒ slope = 3/2

B) y = 3/2x + 9/2 ⇒ slope = 3/2

C) y= 2/3x + 1/3 ⇒ slope = 2/3

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What is the equation of the line that passes through the points (5, 3) and (-3,-1)?
Liono4ka [1.6K]

Answer:

y=1/2x+1/2

m=1/2

Step-by-step explanation:

You want to find the equation for a line that passes through the two points:

(5,3) and (-3,-1).

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (5,3), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=5 and y1=3.

Also, let's call the second point you gave, (-3,-1), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=-3 and y2=-1.

Now, just plug the numbers into the formula for m above, like this:

m=

-1 - 3 over

-3 - 5

or...

m=

-4 over

-8

or...

m=1/2

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=1/2x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(5,3). When x of the line is 5, y of the line must be 3.

(-3,-1). When x of the line is -3, y of the line must be -1.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=1/2x+b. b is what we want, the 1/2 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (5,3) and (-3,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(5,3). y=mx+b or 3=1/2 × 5+b, or solving for b: b=3-(1/2)(5). b=1/2.

(-3,-1). y=mx+b or -1=1/2 × -3+b, or solving for b: b=-1-(1/2)(-3). b=1/2.

See! In both cases we got the same value for b. And this completes our problem.

The equation of the line that passes through the points

(5,3) and (-3,-1)

is

y=1/2x+1/2

7 0
3 years ago
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