Question

Establish the identity.

(cot u + csc u -1)
__________________ = csc u + cot u
(cot u - csc u +1)


I've gotten it to:
((cos u - sin u)+1)
__________________
((cos u + sin u) -1)

That is where I get stuck. :/

Answer 1

Do you want me to proceed from the actual question or the step where you have stuck now, by the way both are one and the same thing..

Answer 2

The technique is to use Rationalization Method here.. It means Multiplying and Dividing by the Denominator by changing a sign to simplify the expression easily, as for example here:

Denominator is : \(cot(u) - cosec(u) + 1\), so you should multiply and divide by \(cot(u) - cosec(u) \color{red}{-} 1\) (See, I have changed +1 to -1..

Now going further:

\[\frac{\cot(u) + cosec(u) - 1}{\cot(u) - cosec(u) +1} \times \frac{\cot(u) - cosec(u) - 1}{\cot(u) - cosec(u) -1}\]

\[\implies \frac{[(\cot(u) - 1)^2 - cosec^2(u)]}{[(\cot(u) - cosec(u))^2 - 1]}\]
\[\implies \frac{\cot^2(u) + 1 - 2\cot(u) + cosec^2(u)}{\cot^2(u) + cosec^2(u) - 2\cot(u)cosec(u) - 1}\]
\[\implies \frac{\cancel{cosec^2(u)} - 2 \cot(u) - \cancel{cosec^2(u)}}{\cot^2(u) + \cancel{1} + \cot^2(u) - 2\cot(u)cosec(u) - \cancel{1}}\]
\[\implies \frac{-2 \cot(u)}{2 \cot^2(u) - 2 \cot(u) cosec(u)} \implies \frac{\cancel{-2 \cot(u)}}{\cancel{-2\cot(u)}[cosec(u) - \cot(u)]}\]

\[\color{green}{\implies \frac{1}{cosec(u) - \cot(u)}}\]

Answer 3

now, you are almost there in front of answer, just Rationalize it once more with \(cosec(u) + cot(u)\), just one step you are away from getting there.. :)

Answer 4

Identities that I have used here is/are:

\[a) \color{red}{\; \; 1 + \cot^2(u) = cosec^2(u)}\]
\[b) \; \; \color{green}{cosec^2(u) - \cot^2(u) = 1}\]

And these are one and the same thing..

Answer 5

A similar type of question you will find at : http://openstudy.com/study#/updates/516cef8fe4b063fcbbbcbf77

You must see it and try to solve it yourself...

Answer 6

Thank you very much! I have already solved this problem at this point, I forgot to come back here and take this down. But your explanation of what is going on helps SO much! Thank you again for taking the time to explain this. :)

Answer 7

It is okay @EagleCore... You are welcome dear..