Verifying Trig Identity: csc(x) + sec(x) = cot(x) + tan(x)

  • Thread starter Trizz
  • Start date
In summary, the given identity can be verified by changing all trigonometric functions to sine and cosine, factoring out a common term in the denominator, canceling common factors, and then using the identity sin^2x + cos^2x = 1 to simplify the fraction.
  • #1
Trizz
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Homework Statement


Verify the identity:

csc(x) + sec(x) = cot(x) + tan(x)
sin(x) + cos(x)


Homework Equations



sin^2 + cos^2 = 1
tan^2 + 1 = sec^2
1 + cot^2 = csc^2



The Attempt at a Solution



First i changed everything to sin and cos to try and make it more clear. Let me know if I should bypass this:

1/sin(x) + 1/cos(x) = cot(x) + tan(x)
sin(x) + cos(x)

I'm not really sure where to go from here...should i try to make a common denominator?
 
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  • #2
Multiply the numerator and denominator by sin(x)cos(x)
 
  • #3
Thanks rock.freak!

Ok i did that and got:

cos(x) + sin(x)
sin(x)^2 + cos(x)^2


does that look right?
 
  • #4
Or is it this?

sin(x) + cos(x)
sin(x)^2cos(x) + sin(x)cos(x)^2
 
  • #5
Trizz said:
Or is it this?

sin(x) + cos(x)
sin(x)^2cos(x) + sin(x)cos(x)^2

It's this one. The other is obviously not true since [itex]sin^2x+cos^2x=1[/itex] so the denominator would become 1.

Don't expand out the denominator, instead, see if you can cancel any common factors.
After this step, it should be easier to work on the right-hand side to try get the left-hand side.
 
  • #6
Thanks Mentallic.

So I kind of looked at the denominator like this (a^2*b)(a*b^2).

So I should be able to pull out one "a" and one "b", right?

Does this seem right?

sin(x) + cos(x)

sin(x)cos(x)(sin(x)^2 + cos(x)^2)
 
  • #7
Actually the denominator is more like [itex]a^2b+b^2a[/itex].

Now you can factor out [itex]ab[/itex] to get [itex]ab(a+b)[/itex]. Do you see how this works?

For the denominator it should be [itex]sinx.cosx\left(sinx+cosx\right)[/itex] so do you see what can be canceled in that fraction?
 
  • #8
Thanks a lot Mentallic.

Yes, I see what you did on the factoring. I did that too quickly.

So with the new denominator, it would give me:

sin(x) + cos(x)
sin(x).cos(x) (sin(x) +cos(x))


It seems like you can cancel out the sin(x) + cos(x) of the numerator with the same thing in parenthesis in the denominator, but am I wrong? Can't you only cancel out things that are multiplied or divided, not attached through addition?

If so, I'm lost on what to cancel out.
 
  • #9
It looks like there was no need to distribute sinxcosx in the denominator at the beginning; just better to have left it like (sinx + cosx)sinxcosx.

Yes, sinx + cosx will cancel, then you have 1/(sinx*cosx). Then use the identity 1 = sin2x + cos2x for the numerator.
 
  • #10
Thanks Bohrok.

I'm still a little confused about how that would verify the identity

Are you saying to set up the fraction like this now?

sin^2x + cos^2x
sinx*cosx

I see the exponent and denominator would cancel out, but how would that give me cot(x) + tan(x) as the final solutions
 
  • #11
Trizz said:
Thanks a lot Mentallic.

Yes, I see what you did on the factoring. I did that too quickly.

So with the new denominator, it would give me:

sin(x) + cos(x)
sin(x).cos(x) (sin(x) +cos(x))


It seems like you can cancel out the sin(x) + cos(x) of the numerator with the same thing in parenthesis in the denominator, but am I wrong? Can't you only cancel out things that are multiplied or divided, not attached through addition?

If so, I'm lost on what to cancel out.

Yes you're allowed to cancel the [itex]sinx+cosx[/itex] from the numerator and denominator. They're both the same factor.

e.g. if we have a fraction such as:

[tex]\frac{xy}{x}[/tex]
we can cancel x from both the numerator and denominator because they're common factors.
This rules works for whatever x may stand for, such as something more complicated like [itex]sina+cosb[/itex]
Then the fraction would be: [tex]\frac{(sina+cosb)y}{sina+cosb}[/tex] and now you can cancel in the same fashion.


Ok so after you cancel the [itex]sinx+cosx[/itex] you will have [tex]\frac{1}{sinx.cosx}[/tex].
It is a little hard to see what you could do with this to try get the RHS, so it would be easier to work on the RHS from this point.
 
  • #12
Trizz said:
Are you saying to set up the fraction like this now?

sin^2x + cos^2x
sinx*cosx

I see the exponent and denominator would cancel out, but how would that give me cot(x) + tan(x) as the final solutions

Then simplify the fraction using the fact that (a + b)/c = a/c + b/c. You could work on the right hand side as Mentallic said so you can see how it should work out.
 
  • #13
Thank you guys for being so kind and giving me your time! I think I can get the rest from here.
 
  • #14
No problem, good luck!
 

1. What is a trigonometric identity?

A trigonometric identity is a mathematical equation that is true for all values of the variables involved. These identities are used to simplify or verify complex trigonometric expressions.

2. How do you verify a trigonometric identity?

To verify a trigonometric identity, you need to manipulate the expression using basic trigonometric identities and algebraic techniques to show that both sides of the equation are equal.

3. What are some common trigonometric identities?

Some common trigonometric identities include the Pythagorean identities, double angle identities, half angle identities, and sum and difference identities.

4. Why is it important to verify trigonometric identities?

Verifying trigonometric identities is important because it helps in simplifying complex expressions and solving equations involving trigonometric functions. It also helps in understanding the relationships between different trigonometric functions.

5. What are some tips for verifying trigonometric identities?

Some tips for verifying trigonometric identities include starting with the more complex side of the equation, using basic identities and algebraic techniques, and checking for any restrictions on the variables involved.

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