Question

Prove this identity:

tan(A)/(1-cot(A)) + cot(A)/(1-tan(A)) = sec(A)*cosec(A)+1

Answer 1

Hey welcome to OpenStudy!\[\large\rm \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\sec A \csc A +1\]
This is the equation we're trying to prove?
Does that look correct?

Answer 2

Try to turn one side into the other side. make the left side become sec(A)csc(A)+1 to show that both sides are the same

Answer 3

@zepdrix Yes that is what I want to prove. Please give a step by step solution, thank you.

Answer 4

lots of algebra to turn the left side into the right side...

Answer 5

ill help ya if you need
\[\frac{ \tan(A) }{ 1-\frac{ 1 }{ \tan(A) } }+\frac{ \frac{ 1 }{ \tan(A) } }{ 1-\tan(A) }=\sec(A)
\csc(A)+1\]

Answer 6

@DanJS interesting lead.

Answer 7

need the same denominators to add those on the left side.. you can multiply the first denominator by tan(A)/tan(A), the second term on the left just moves the tan(A) to the bottom, ill just keep putting only the left side

\[\frac{ \tan(A) }{ \frac{ \tan(A)-1 }{ \tan(A) } }+\frac{ 1 }{\tan(A)( 1-\tan(A)) }=\]
right?

Answer 8

that goes to, after simplifying the first term

\[\large \frac{ \tan^2(A) }{ \tan(A)-1 }+\frac{ 1 }{ \tan(A)(1-\tan(A)) }\]

you good on that so far?

Answer 9

Thanks a lot for the help @DanJS I was able to solve the rest. :)

Answer 10

ok, welcome, when i did it there was a cubic expansion for tan^3(A)-1

Answer 11

Yes I encountered the cubic expansion. I used the formula (a-b)(a^2+ab+b^2). Thanks a lot for your help.

Answer 12

welcome

Answer 13

What about this?