Question

Prove that
cosecA/cosecA - 1 + cosec A/cosecA + 1 = 2sec^2A
@welshfella
@loser66

Answer 1

@TheSmartOne

Answer 2

@perl

Answer 3

@jim_thompson5910

Answer 4

@zepdrix

Answer 5

Ah, I'm not able to think properly right now and I'm tired. :/

Maybe @zepdrix @Nnesha @jim_thompson5910 @perl can help you better :)

Answer 6

Is this copied correctly?
$$
\Large
\frac{\csc A}{\csc(A )- 1} + \frac{\csc A}{\csc(A )+ 1} = 2\sec^2A

$$

Answer 7

Yes, @perl @zepdrix and no @Nnesha

Answer 8

Now you're correct
@Nnesha

Answer 9

You can start with either side, lets go with the left side since we can simplify that
$$\large{
\frac{\csc A}{\csc(A )- 1} + \frac{\csc A}{\csc(A )+ 1}
\\ ~\\= \color{red}{\frac{\csc (A)-1}{\csc(A )- 1}}\cdot \frac{\csc A}{\csc(A )- 1} + \frac{\csc A}{\csc(A )+ 1}\cdot \color{red}{\frac{\csc (A)+1}{\csc(A )+ 1}}
}$$

Answer 10

$$\large {
\frac{\csc A}{\csc(A )- 1} + \frac{\csc A}{\csc(A )+ 1}
\\ ~\\= \normalsize
\left(\color{red}{\frac{\csc (A)-1}{\csc(A )- 1}}\right)
\cdot \left(\frac{\csc A}{\csc(A )- 1}\right) + \left(\frac{\csc A}{\csc(A )+ 1}\right)
\cdot \left(\color{red}{\frac{\csc (A)+1}{\csc(A )+ 1}} \right)
}$$

Answer 11

The thing is not clear I mean I can't see the latex Its like I see half only.

Answer 12

can you see it now?

Answer 13

Nope :(

Answer 14

you can try refreshing

Answer 15

I did but It ain't working :/

Answer 16

youre saying your latex does not show?

Answer 17

can you take a screenshot of it

Answer 18

http://prntscr.com/75dbgk

Answer 19

thats weird

Answer 20

this is what i see http://prntscr.com/75dblz

Answer 21

anyways, its not a problem :)

Answer 22

Can you type it out please?

Answer 23

Hmm, seems like you are using a smaller screen thus the issues.

Answer 24

$${
\frac{\csc A}{\csc(A )- 1} + \frac{\csc A}{\csc(A )+ 1}
\\ ~\\=
\left(\color{red}{\frac{\csc (A)-1}{\csc(A )- 1}}\right)
\cdot \left(\frac{\csc A}{\csc(A )- 1}\right)
\\+ \left(\frac{\csc A}{\csc(A )+ 1}\right)
\cdot \left(\color{red}{\frac{\csc (A)+1}{\csc(A )+ 1}} \right)
}$$

Answer 25

is that better?

Answer 26

YES!

Answer 27

ok great :)

Answer 28

note that what i multiplied by was not arbitrary, we want a common denominator

Answer 29

and its a valid multiplication, since it is equivalent to multiplying by 1. therefore i did not change the expression

Answer 30

But I was told to do this

cosec A/cosecA-1 x cosec A +1/cosec A + 1

Answer 31

thats essentially the same thing, just skipping the step of common denominator

Answer 32

$$ \Large{ \frac a b + \frac c d
\\~\\ = \frac{a\cdot d + b\cdot c }{b\cdot d}


}$$

Answer 33

What next?

Answer 34

$$ \Large{ \frac A B + \frac C D = \frac{AD + BC }{BD}
}$$

Answer 35

???

Answer 36

do you agree with that rule

Answer 37

Sure.

Answer 38

$$
{
\frac{\csc A}{\csc(A )- 1} + \frac{\csc A}{\csc(A )+ 1}
\\ ~\\
= \frac{\csc A \cdot ( \csc(A )+ 1) + \csc A \cdot (\csc(A )- 1) }{(\csc(A )- 1)~(\csc(A )+ 1)}

}$$

Answer 39

???

Answer 40

Is it like
2csc^2A/cot^2A?

Answer 41

$$
{
\frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1}
\\ ~\\
= \frac{\csc A \cdot ( \csc A+ 1) + \csc A \cdot (\csc A - 1) }{(\csc A - 1)~(\csc A + 1)}
\\~\\
= \frac{\csc^2 A + \csc A + \csc^2 A- \csc A }{(\csc A - 1)~(\csc A + 1)}
\\~\\
= \frac{2\csc^2 A }{(\csc A - 1)~(\csc A + 1)}
\\~\\
= \frac{2\csc^2 A }{\csc^2 A - \csc A + \csc A - 1}
\\~\\
= \frac{2\csc^2 A }{\csc^2 A - 1}

}$$

Answer 42

csc^2A - 1 = cot^2A?

Answer 43

right :)

Answer 44

$$
{
\frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1}
\\ ~\\
= \frac{\csc A \cdot ( \csc A+ 1) + \csc A \cdot (\csc A - 1) }{(\csc A - 1)~(\csc A + 1)}
\\~\\
= \frac{\csc^2 A + \csc A + \csc^2 A- \csc A }{(\csc A - 1)~(\csc A + 1)}
\\~\\
= \frac{2\csc^2 A }{(\csc A - 1)~(\csc A + 1)}
\\~\\
= \frac{2\csc^2 A }{\csc^2 A - \csc A + \csc A - 1}
\\~\\
= \frac{2\csc^2 A }{\csc^2 A - 1}
\\~\\
= \frac{2\csc^2 A }{\cot^2 A }
\\~\\ =\Large 2\cdot \frac{\frac{1}{\sin ^2 A } }{\frac 1 {1\tan^2 A} }
\\~\\ =\large 2\cdot \frac{1}{\sin^2 A } \cdot \frac{\tan^2 A }{1}
\\~\\ =\large 2\cdot \frac{1}{\sin^2 A } \cdot \frac{\sin^2 A }{\cos^2 A }
\\~\\ =\large 2\cdot \frac{1}{\cancel \sin^2 A } \cdot \frac{\cancel \sin^2 A }{\cos^2 A }
\\~\\ =\large 2\cdot \frac{1}{\cos^2 A }
\\~\\ =\large 2\cdot \sec^2 A
}$$

Answer 45

Thanks,
can u help me more?