Question

Verify [(cot alpha/sec alpha- tan alpha)-(cos alpha/sec alpha +tan alpha)]= sin alpha+csc alpha

Answer 1

Hmm let's see if we can get this formatted correctly,\[\Large\rm \frac{\cot \alpha}{\sec \alpha- \tan \alpha}-\frac{\cos \alpha}{\sec \alpha+\tan \alpha}=\sin \alpha+\csc \alpha\]Does that look close maybe?

Answer 2

Yes, it does/

Answer 3

These identities can be tough sometimes.
There are so many ways to approach them.

May first instinct is to recognize that the denominators are conjugates of one another, so maybe we can get a common denominator.
Hmm thinking whether that's the way we want to go or not...

Answer 4

So, multiply the LHS fractions by their conjugate?

Answer 5

No that path seems bad.
Kinda makes the problem messier.
Let's try this instead,\[\Large\rm \frac{\cot \alpha}{\color{orangered}{\sec \alpha- \tan \alpha}}-\frac{\cos \alpha}{\color{orangered}{\sec \alpha+ \tan \alpha}}=\sin \alpha+\csc \alpha\]Let's write these orange parts in terms of sines and cosines.

Answer 6

My first instinct was wrong hehe.
So anyway, sines and cosines gives us,\[\Large\rm \frac{\cot \alpha}{\color{orangered}{\frac{1}{\cos \alpha}-\frac{\sin \alpha}{\cos \alpha}}}-\frac{\cos \alpha}{\color{orangered}{\sec \alpha+ \tan \alpha}}=\sin \alpha+\csc \alpha\]No no no no no.
Ok ok the conjugate thing would have been a good idea actually :P

Answer 7

Ok sorry sorry I keep flipping back and forth lol.

Answer 8

No, you're fine. I really appreciate the help!

Answer 9

Would you also write the second denominator in terms of sines and cosines?

Answer 10

Ah yes ok fine we'll stick with that path ^^ hehe.
Yes the other denominator as well.

Answer 11

Whichever path you prefer is fine.

Answer 12

\[\Large\rm \frac{\cot \alpha}{\color{orangered}{\frac{1}{\cos \alpha}-\frac{\sin \alpha}{\cos \alpha}}}-\frac{\cos \alpha}{\color{orangered}{\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}}}=\sin \alpha+\csc \alpha\]
\[\Large\rm \frac{\cot \alpha}{\color{orangered}{\frac{1-\sin \alpha}{\cos \alpha}}}-\frac{\cos \alpha}{\color{orangered}{\frac{1+\sin \alpha}{\cos \alpha}}}=\sin \alpha+\csc \alpha\]See how both terms have a cosine in the bottom?
So we can combine those fractions.

Answer 13

Oh yes yes this path is going to be bad... oh boy.. let's not keep going down this :C
My bad my bad.

Answer 14

Back tot he cojugates.

Answer 15

It's fine. We can go throught the conjugate path.

Answer 16

\[\large\rm \frac{\cot \alpha\color{royalblue}{(\sec \alpha+\tan \alpha)}}{(\sec \alpha- \tan \alpha)\color{royalblue}{(\sec \alpha+\tan \alpha)}}-\frac{\cos \alpha\color{#CC0033}{(\sec \alpha-\tan \alpha)}}{(\sec \alpha+\tan \alpha)\color{#CC0033}{(\sec \alpha-\tan \alpha)}}\]

Answer 17

So our denominators become \(\Large\rm \sec^2\alpha-\tan^2\alpha\)

Answer 18

Let's look back at our Pythagorean Identity and see if we can do anything with that.

Answer 19

\[\Large\rm 1+\tan^2x=\sec^2x\]Can we use this to simplify our denominator maybe?

Answer 20

Okay, that works.

Answer 21

So what does our denominator simplify to? ;)

Answer 22

Comeon lyss! You can do this step!
I know you can! :3

Answer 23

Wouldn't that equal positive one? At least with sec^2-tan^2.

Answer 24

Yes good.

Answer 25

\[\Large\rm \frac{\cot \alpha(\sec \alpha+\tan \alpha)}{1}-\frac{\cos \alpha(\sec \alpha-\tan \alpha)}{1}\]

Answer 26

So that deals with the fractions entirely, which is nice.\[\Large\rm \cot \alpha(\sec \alpha+\tan \alpha)-\cos(\sec \alpha-\tan \alpha)\]

Answer 27

So I guess we need to expand these out.
It might be a good idea to convert to sines and cosines at this point.
Or maybe after the distributing, I dunno.

Answer 28

\[\Large\rm \cot \alpha \sec \alpha + \cot \alpha \tan \alpha- \cos \alpha \sec \alpha+ \cos \alpha \tan \alpha\]Oh boy :d

Answer 29

Look at these two middle terms a sec,
\[\Large\rm \cot \alpha \sec \alpha \color{orangered}{+ \cot \alpha \tan \alpha- \cos \alpha \sec \alpha}+ \cos \alpha \tan \alpha\]The first one should simply down really nicely.
Remember how to convert cotangent and tangent to sines and cosines?

Answer 30

Yes, \[(\cos \alpha/ \sin \alpha)(\sin \alpha/\cos \alpha)\]

Answer 31

Mmm k, and does that simplfy? :)

Answer 32

\[(1/\sin \alpha)(\sin \alpha0\]

Answer 33

the zero wasn't suppose to be in there. Sorry.

Answer 34

Ok good.
It should simplify further though.
The cosines cancelled out.
The sines can cancel out as well, yes? (One is in top-ish, the other in the bottom).

Answer 35

Yeah. I was wondering if they also cancelled out.

Answer 36

So, all we're left with is cot alpha*sec alpha+cos alpha*tan alpha?

Answer 37

Ok so you were able to work ahead a little bit?
Each orange term gives you 1, they add to 0.
Good good good.

Answer 38

Yup!
Just a tiny bit of simplification to wrap this one up!

Answer 39

Alright, I believe I have it from here. Thank you!!

Answer 40

Yay team!