# How should I have solved this?

• Jan 19th 2011, 07:45 AM
NathanBUK
How should I have solved this?
So just got my exam and there was this question on it that I just didn't know where to start, wondered if any of you could explain to me how it should have been done.

Had to show that $(cosecx)/(cosecx+1)+(cosecx)/(cosecx-1)=50$ could be written as $sec^2x=25$

I'm pretty sure that was the question, if its not possible then the + is probably a -after the first fraction.

Also the next part was to solve $sec^2x = 25$ and all my friends got 2 answers and I got 4, does that sound right?

Only identies we can use in this exam are

$tan^2x+1=sec^2x$
$cot^2x+1=cosec^2x$
$sin^2x+cos^2x=1$
and all the 1 overs for tan sin and cos.

Thanks!
• Jan 19th 2011, 07:55 AM
e^(i*pi)
Throughout I will use $\csc(x) = cosec(x)$ because I'm lazy

A good start will be to get a common denominator which will be $(\csc(x)+1)(\csc(x)-1)$.

$\dfrac{\csc(x)(\csc(x)-1)}{(\csc(x)+1)} + \dfrac{\csc(x)(\csc(x)+1)}{\csc(x)-1} = \dfrac{\csc(x)(\csc(x)-1) + \csc(x)(\csc(x)+1)}{(\csc(x)-1)(\csc(x)+1)}$

When you expand you get (note that the denominator is difference of two squares)

$\dfrac{\csc^2(x)-1 + \csc^2(x)+1}{\csc^2(x)-1} = \dfrac{2\csc^2(x)}{\csc^2(x)-1}$

Spoiler:
Using the identities given we get $\dfrac{2\csc^2(x)}{\cot^2(x)} = \dfrac{2}{\sin^2(x)} \cdot \dfrac{\sin^2(x)}{\cos^2(x)}$

since we're dividing by a fraction I flipped it and multiplied. Cancelling gives

$\dfrac{2}{\cos^2(x)} = 2\sec^2(x)$

Set this equal to 50 and divide by 2 to finish the question

edit: don't want to give the whole working away so put some in a spoiler. Look if you want, tis no water off my back