Trig Identity

[Rsls79]

Can anyone start this. I've tried to start with the right and left side separately and the answer look nothing like the other side.



would you start by making cot and tan into sin & cos or would you only make cot 1/tan?

 

[soroban]

Hello, Rsls79!

\(\displaystyle \L\frac{\cot\theta}{1\,-\,\tan\theta}\,+\,\frac{\tan\theta}{1\,-\,\cot\theta}\:\)\(\displaystyle =\:\cot\theta\,+\,\tan\theta\,+\,1\)

Both your ideas are good and will work.
\(\displaystyle \;\;\)But either way, it's a very messy problem!

The left side is: \(\displaystyle \L\,\frac{\frac{1}{\tan\theta}}{1\,-\,\tan\theta}\,+\,\frac{\tan\theta}{1\,-\,\frac{1}{\tan\theta}}\)


Multiply top and bottom of both fractions by \(\displaystyle \tan\theta:\)

\(\displaystyle \L\;\;\;\frac{1}{\tan\theta(1\,-\,\tan\theta)}\,+\,\frac{\tan^2\theta}{\tan\theta\,-\,1} \;= \;\frac{1}{\tan\theta(1\,-\,\tan\theta)}\,-\,\frac{\tan^2\theta}{1\,-\,\tan\theta}\)


Get a common denominator: \(\displaystyle \L\:\frac{1}{\tan\theta(1\,-\,\tan\theta)} \,- \,\frac{\tan^3\theta}{\tan\theta(1\,-\,\tan\theta)}\)

Combine: \(\displaystyle \L\:\frac{1\,-\,\tan^3\theta}{\tan\theta(1\,-\,\tan\theta)}\)

The numerator is a difference of cubes: \(\displaystyle \L\:\frac{(1\,-\,\tan\theta)(1\,+\,\tan\theta\,+\,\tan^2\theta)}{\tan\theta(1\,-\,\tan\theta)}\)


Can you finish it now?

 

[Rsls79]

oooooh! Thank you. I sure can. Was doing so many things wrong thanks!!