Jan 24, 2010
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What is the area of the figure bounded by the line x=4, x axis, and line y=x ?

The answer explanation assumes that the figure formed by these 3 lines is a triangle but im seeing a square. Can any one explain how these lines form a triangle and not a square. The answer is 8 btw.

One more question. How do you solve this?

Tan(x) +cot(x) / csc(x)

the answer is sec(x)

thanks!!
 
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Maygyver

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The line x=4 will be a vertical line at x=4. The line y=x will be a diagonal line bisecting the top right quadrant. The x axis is a horizontal line at y=0. Does that make sense?
 

Streetwolf

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What is the area of the figure bounded by the line x=4, x axis, and line y=x ?

The answer explanation assumes that the figure formed by these 3 lines is a triangle but im seeing a square. Can any one explain how these lines form a triangle and not a square. The answer is 8 btw.

One more question. How do you solve this?

Tan(x) +cot(x) / csc(x)

the answer is sec(x)

thanks!!
1. Draw those lines as if you were individually plotting the equations... however, put them on the same graph.

2. tan(x) + cot(x)/csc(x)

Since order of operations doesn't work here, I'm going to assume you left out some parentheses and do tan(x) + cot(x) first.

tan(x) + cot(x) = sin(x)/cos(x) + cos(x)/sin(x)

= sin^2(x)/sin(x)cos(x) + cos^2(x)/sin(x)cos(x)

= sin^2(x) + cos^2(x) / sin(x)cos(x)

= 1/sin(x)cos(x)

Then divide by csc(x) = 1/sin(x)

= 1/sin(x)cos(x) divided by 1/sin(x)

= 1/cos(x)

= sec(x)

============

Option 2:

tan(x) + cot(x) all over csc(x) can be written as tan(x)/csc(x) + cot(x)/csc(x)

= (sin/cos) / (1/sin) + (cos/sin) / (1/sin)

= sin^2/cos + cos

= sin^2/cos + cos^2/cos

= sin^2 + cos^2 all over cos

= 1/cos

= sec(x)

I just left all the (x) out of the 2nd way... they'd still be there.
 

evanyou

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What is the area of the figure bounded by the line x=4, x axis, and line y=x ?

The answer explanation assumes that the figure formed by these 3 lines is a triangle but im seeing a square. Can any one explain how these lines form a triangle and not a square. The answer is 8 btw.

One more question. How do you solve this?

Tan(x) +cot(x) / csc(x)

the answer is sec(x)

thanks!!

Here's how I deal with tirg

You remember SOH CAH TOA right?

Tan(x) is O/A and cot(x) is A/O.

The sum of them would be (O^2+A^2)/O*A which is H^2/O*A.

now divide it by csc(x) which is H/O (reciprocal function of sin (OH))

it finally gives H/A which is Sec(x)

you might want to use x or y instead of A or O :idea:
 
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How is the line y=x a diagonal line bisecting the right quadrant? Is that a the deifnition of y=x? I took it as saying y=x which means y=4 becuase x=4?
 
Jan 24, 2010
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And also replying to evanyou. How did you know that ''The sum of them would be (O^2+A^2)/O*A which is H^2/O*A.''

Where did h^2 come from?
 

Maygyver

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How is the line y=x a diagonal line bisecting the right quadrant? Is that a the deifnition of y=x? I took it as saying y=x which means y=4 becuase x=4?
Just think of the definition of a line. y=mx+b so y=x is just a line with a slope of 1
 

Streetwolf

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And also replying to evanyou. How did you know that ''The sum of them would be (O^2+A^2)/O*A which is H^2/O*A.''

Where did h^2 come from?
Pythagorean theorem.