Trigonometric identity?

Question

I need help on a few identity problems that I consider EXTREMELYchallenging and confusing. Here they are:

Note: I used the letter "O" in replacement for theta.

> (cotO / 1 - tanO) + (tanO / 1- cotO) = 1 + tanO + cotO
> sinOcosO / cos²O-sin²O = tanO / (1-tan²O)
> sinO-cosO+1 / sinO+cosO-1 = sinO+1 / cosO
> secO-cosO / secO+cosO = sin²O / 1 + cosO
> (tanO-cotO) / (tanO+cotO) + 2 cos²O = 1
> secO / 1 + secO = 1 - cosO / sin²O

Thanks for your help!

For the third question, it is supposed to be: secO-cosO / secO+cosO = sin²O / 1 + cos²O, not secO-cosO / secO+cosO = sin²O / 1 + cosO

Answer 1

The general method of solving trig identities is as follows:
1) Choose the more complex side and work with that
2) Change everything to sines and cosines
3) Get rid of all complex fractions
4) Try to make it look like the other side.

If those 4 steps don't work, stop your solution right there and work with the other side. Try to get the other side to look like your stopping point.

I'll solve one of them for you using those guidelines. I'm going to use a "t" instead of an O.

cot(t) / [1 - tan(t)] + tan(t) / [1 - cot(t)] = 1 + tan(t) + cot(t)

LHS = cot(t) / [1 - tan(t)] + tan(t) / [1 - cot(t)]

Change everything to sines and cosines.

LHS = [cos(t)/sin(t)] / [1 - sin(t)/cos(t)] + [sin(t)/cos(t)] [1 - cos(t)/sin(t)]

Get rid of all complex fractions (i.e. fractions within fractions). I'm going to multiply the first expression by sin(t)cos(t) on the numerator and denominator, and the same thing with the second expression.

LHS = [cos^2(t)] / [sin(t)cos(t) - sin^2(t)] + [sin^2(t)] / [sin(t)cos(t) - cos^2(t)]

Now, let's factor the denominators accordingly.

LHS = [cos^2(t)] / [sin(t) (cos(t) - sin(t))] + [sin^2(t)] / [cos(t) (sin(t) - cos(t)]

All we can do now is merge these two fractions into one. Let's give them a common denominator of cos(t)sin(t) (cos(t) - sin(t)) But to do that, we'd have to factor -1 out of the second denominator to change cos(t) - sin(t) into sin(t) - cos(t). Doing so, we have

LHS = [cos^2(t)] / [sin(t) (cos(t) - sin(t))] - [sin^2(t)] / [cos(t) (cos(t) - sin(t)]

Making a common denominator, we have

LHS = [cos^3(t) - sin^3(t)] / [sin(t) cos(t) (cos(t) - sin(t))]

Now, we can factor the numerator as a difference of cubes.

LHS = [(cos(t) - sin(t)) (cos^2(t) + cos(t)sin(t) + sin^2(t))] /
[sin(t) cos(t) (cos(t) - sin(t))]

Note the cancellation of cos(t) - sin(t), leaving us with

LHS = [cos^2(t) + cos(t)sin(t) + sin^2(t)] / (sin(t)cos(t))

Splitting this up into three fractions, we get

LHS = [cos^2(t)]/[sin(t)cos(t)] + [cos(t)sin(t)] / [cos(t)sin(t)]
+ [sin^2(t)] / [sin(t)cos(t)]

And a LOT of cancelling takes places.

LHS = cos(t)/sin(t) + 1 + sin(t)/cos(t)

By definition, these are equal to

LHS = cot(t) + 1 + tan(t)

Which we can rearrange because addition is commutative.

LHS = 1 + tan(t) + cot(t) = RHS

***
2) sin(t)cos(t) / [cos^2(t) - sin^2(t)] = tan(t) / [1 - tan^2(t)]

RHS = tan(t) / [1 - tan^2(t)]

RHS = [sin(t)/cos(t)] / [1 - (sin(t)/cos(t))^2]

Note that squaring a fraction means squaring the top and bottom.

RHS = [sin(t)/cos(t)] / [1 - (sin^2(t) / cos^2(t))]

Multiply top and bottom by cos^2(t) to get

RHS = [sin(t)cos(t)] / [cos^2(t) - sin^2(t)] = LHS

3) [sin(t) - cos(t) + 1] / [sin(t) + cos(t) - 1] = [sin(t) + 1] / cos(t)

LHS = [sin(t) - cos(t) + 1] / [sin(t) + cos(t) - 1]

This is a difficult identity, considering everything is already in sines and cosines. That's why we have to try a few tricks.
Let's multiply numerator and denominator by the conjugate of the denominator; that is [sin(t) + cos(t) + 1]

LHS = [(sin(t) - cos(t)) + 1][(sin(t) + cos(t)) + 1] /
[(sin(t) + cos(t)) - 1] [(sin(t) + cos(t)) + 1]

This will give us a difference of squares in the denominator. We can also expand the numerator.

LHS = [ (sin(t) - cos(t)) (sin(t) + cos(t)) + sin(t) - cos(t) + sin(t) + cos(t) + 1 ] / [ (sin(t) + cos(t))^2 - 1 ]

We can simplify the top; some terms cancel out, even.

LHS = [ sin^2(t) - cos^2(t) + 2sin(t) + 1 ] / [sin^2(t) + 2sin(t)cos(t) + cos^2(t) - 1]

From the denominator, we can see a sin^2(t) + cos^2(t). We can make that into 1.

LHS = [ sin^2(t) - cos^2(t) + 2sin(t) + 1 ] / [1 + 2sin(t)cos(t) - 1]

And notice the 1 and the -1 will obliterate each other.

LHS = [ sin^2(t) - cos^2(t) + 2sin(t) + 1 ] / [2sin(t)cos(t)]

In the numerator, notice we can make out a 1 - cos^2(t). We can change this to sin^2(t).

LHS = [ sin^2(t) + sin^2(t) + 2sin(t) ] / [2sin(t)cos(t)]

Grouping like terms,

LHS = [ 2sin^2(t) + 2sin(t) ] / [2sin(t)cos(t)]

All in one step, I'm going to factor out a 2sin(t) on the top, which will cancel out with the 2sin(t) on the bottom, and leave us with

LHS = [ sin(t) + 1 ] / [cos(t)] = RHS

***
4) [sec(t) - cos(t)] / [sec(t) + cos(t)] = sin^2(t) / [1 + cos(t)]

LHS = [sec(t) - cos(t)] / [sec(t) + cos(t)]

Change everything to sines and cosines (if applicable)

LHS = [1/cos(t) - cos(t)] / [1/cos(t) + cos(t)]

Multiply top and bottom by cos(t), to eliminate the complex fraction.

LHS = [1 - cos^2(t)] / [1 + cos^2(t)]

LHS = sin^2(t) / [1 + cos^2(t)]

Are you sure you wrote down the identity correctly? There's no way sin^2(t) / [1 + cos^2(t)] will equal what is written there (unless I'm missing something). I'm going to assume it's a typo.

***
5) [tan(t) - cot(t)] / [tan(t) + cot(t)] + 2cos^2(t) = 1

LHS = [tan(t) - cot(t)] / [tan(t) + cot(t)]

LHS = [ sin(t)/cos(t) - cos(t)/sin(t) ] / [sin(t)/cos(t) + cos(t)/sin(t)] + 2cos^2(t)

Multiply top and bottom by cos(t)sin(t), to get

LHS = [ sin^2(t) - cos^2(t) ] / [sin^2(t) + cos^2(t)] + 2cos^2(t)

Note the identity in the denominator is just 1; this leaves us with

LHS = sin^2(t) - cos^2(t) + 2cos^2(t)

Grouping like terms,

LHS = sin^2(t) + cos^2(t) = 1 = RHS

***
6) sec(t) / [1 + sec(t)] = [1 - cos(t)] / sin^2(t)

LHS = sec(t) / [1 + sec(t)]

LHS = (1/cos(t)) / [1 + 1/cos(t)]

Multiply numerator and denominator by cos(t), to get
LHS = 1 / [cos(t) + 1]

Let's rearrange the terms on the bottom, since addition is commutative anyway.

LHS = 1 / [1 + cos(t)]

This is the part that isn't immediately intuitive. What you have to do is multiply the denominator by its conjugate. This means we multiply the numerator by that same conjugate as well, so we multiply top and bottom by (1 - cos(t)). Remember that multiplying (a + b) by a conjugate (a - b) means we get a difference of squares, (a^2 - b^2).

LHS = (1 - cos(t)) / [1 - cos^2(t)]

Note the identity in the denominator ; 1 - cos^2(t) = sin^2(t)

LHS = (1 - cos(t)) / (sin^2(t)) = RHS

Answer 2

I had my trigonometric identity stolen once. It really sucked because I lost sin²O US Dollars and my credit rating turned to cotO.

Answer 3

Tanx + (Cosx)/(a million+Sinx) = Secx artwork on LHS LHS = Tanx + (Cosx)/(a million+Sinx) = sinx/cosx + (Cosx)/(a million+Sinx) = [sinx(a million+Sinx) + Cos²x] / [{cosx}(a million+Sinx)] = [sinx + sin²x + cos²x] / [{cosx}(a million+Sinx)] = [sinx + a million] / [{cosx}(a million+Sinx)] = a million / {cosx} = secx = RHS

Answer 4

thats easy unless you go off on a tangent

Answer 5

(cot(x)/(1 - tan(x)) + (tan(x)/(1 - cot(x))
((1/tan(x))/(1 - tan(x))) + (tan(x)/(1 - (1/tan(x)))
(1/(tan(x)(1 - tan(x))) + (tan(x)/((tan(x) - 1)/(tan(x)))
(1/(tan(x)(1 - tan(x))) + ((tan(x)^2)/(tan(x) - 1))
(1/(tan(x)(1 - tan(x))) + ((tan(x)^2)/(-1 + tan(x)))
(1/(tan(x)(1 - tan(x))) + ((tan(x)^2)/(-(1 - tan(x))))
(1/(tan(x)(1 - tan(x))) - ((tan(x)^2)/(1 - tan(x)))
(1 - tan(x)^3)/(1 - tan(x))
((1 - tan(x))(1 + tan(x) + tan(x)^2))/(tan(x)(1 - tan(x)))
(1 + tan(x) + tan(x)^2)/(tan(x))

(1/tan(x)) + 1 + tan(x) or in our case 1 + tan(x) + cot(x)

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(tan(x))/(1 - tan(x)^2)
(sin(x)/cos(x))/(1 - (sin(x)/cos(x))^2)
(sin(x)/cos(x)) / ((cos(x)^2 - sin(x)^2)/(cos(x)^2)))
(sin(x)/cos(x)) * ((cos(x)^2)/(cos(x)^2 - sin(x)^2)
(sin(x)cos(x)^2)/((cos(x))(cos(x)^2 - sin(x)^2)

(sin(x)cos(x))/(cos(x)^2 - sin(x)^2)

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(sin(x) - cos(x) + 1)/(sin(x) + cos(x) - 1) = (sin(x) + 1)/cos(x)

Multiply top and bottom by (sin(x) - cos(x) + 1)

((sin(x) - cos(x) + 1)^2)/((sin(x) + cos(x) - 1)(sin(x) - cos(x) + 1))

((sin(x) - cos(x) + 1)(sin(x) - cos(x) + 1))/(sin(x)^2 - sin(x)cos(x) + sin(x) + sin(x)cos(x) - cos(x)^2 + cos(x) - sin(x) + cos(x) - 1)

(sin(x)^2 - sin(x)cos(x) + sin(x) - sin(x)cos(x) + cos(x)^2 - cos(x) + sin(x) - cos(x) + 1)/(sin(x)^2 - cos(x)^2 + 2cos(x) - 1)

(sin(x)^2 + cos(x)^2 - 2sin(x)cos(x) + 2sin(x) - 2cos(x) + 1)/(-1 + sin(x)^2 - cos(x)^2 + 2cos(x))

(1 - 2sin(x)cos(x) + 2sin(x) - 2cos(x) + 1)/((-1 + sin(x)^2) - cos(x)^2 + 2cos(x))

(2 - 2sin(x)cos(x) + 2sin(x) - 2cos(x))/(-(1 - sin(x)^2) - cos(x)^2 + 2cos(x))

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

(2 - 2sin(x)cos(x) + 2sin(x) - 2cos(x))/(-2cos(x)^2 + 2cos(x))
(2(1 - sin(x)cos(x) + sin(x) - cos(x))/(-2(cos(x)^2 - cos(x))
-(1 - sin(x)cos(x) + sin(x) - cos(x))/(cos(x)^2 - cos(x))
(-1 + sin(x)cos(x) - sin(x) + cos(x))/(cos(x)(cos(x) - 1))
(sin(x)cos(x) - sin(x) + cos(x) - 1)/(cos(x)(cos(x) - 1))
(sin(x)(cos(x) - 1) + (cos(x) - 1))/(cos(x)(cos(x) - 1)
((cos(x) - 1)(sin(x) + 1))/(cos(x)(cos(x) - 1))

(sin(x) + 1)/(cos(x))

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(sec(x) - cos(x))/(sec(x) + cos(x)) = (sin(x)^2)/(1 + cos(x)^2)

((1/cos(x)) - cos(x))/((1/cos(x)) + cos(x))
((1 - cos(x)^2)/cos(x)) / ((1 + cos(x)^2)/cos(x))
((1 - cos(x)^2)/(cos(x)) * ((cos(x))/(1 + cos(x)^2))
(cos(x)(1 - cos(x)^2))/(cos(x)(1 + cos(x)^2)
(1 - cos(x)^2)/(1 + cos(x)^2)

(sin(x)^2)/(1 + cos(x)^2)

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(tan(x) - cot(x))/(tan(x) + cot(x)) + 2cos(x)^2 = 1

(tan(x) - (1/tan(x)))/(tan(x) + (1/tan(x))) + 2cos(x)^2
(((tan(x)^2 - 1)/(tan(x)))/((tan(x)^2 + 1)/(tan(x))) + 2cos(x)^2
(((tan(x)^2 - 1)/(tan(x)))*((tan(x))/(tan(x)^2 + 1))) + 2cos(x)^2
((tan(x)(tan(x)^2 - 1))/(tan(x)(tan(x)^2 + 1)) + 2cos(x)^2
(tan(x)^2 - 1)/(tan(x)^2 + 1) + 2cos(x)^2
((sin(x)/cos(x))^2 - 1)/((sin(x)/cos(x))^2 + 1) + 2cos(x)^2

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

(((sin(x)^2 - cos(x)^2)/(cos(x)^2))*((cos(x)^2)/(sin(x)^2 + cos(x)^2)) + 2cos(x)^2

(((sin(x)^2 - cos(x)^2)/(cos(x)^2)) * ((cos(x)^2)/1) + 2cos(x)^2
((cos(x)^2)(sin(x)^2 - cos(x)^2))/(cos(x)^2) + 2cos(x)^2
sin(x)^2 - cos(x)^2 + 2cos(x)^2
sin(x)^2 + cos(x)^2

1

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(sec(x))/(1 + sec(x)) = (1 - cos(x))/(sin(x)^2)

(1/cos(x))/(1 + (1/cos(x))
(1/cos(x))/((cos(x) + 1)/cos(x))
(1/cos(x))*((cos(x))/(cos(x) + 1)
(cos(x))/(cos(x)(cos(x) + 1)
1/(cos(x) + 1)

(1 - cos(x))/(sin(x)^2)
(1 - cos(x))/(1 - cos(x)^2)
(1 - cos(x))/((1 - cos(x))(1 + cos(x))

1/(cos(x) + 1)

Sometimes you work the right side to proove the left, sometimes you work the left side to proove the right side, and sometimes you have to work both sides to proove they are equal to each other.